On The Riesz Representation Theorem we played Dodge the Rationals: nested intervals, each one dodging the next rational on a list, closing in on a point that no rational could reach. That proved $\mathbb{Q}$ is not a $G_\delta$. This page plays the same game in every complete metric space at once.
So far, completeness has done exactly one job for us: it makes series converge. Absolutely convergent series converge in a Banach space, and that's how we built Banach spaces in the first place. This page shows completeness doing something much stranger. It makes a complete space big, in a sense that has nothing to do with measure: you can't build one out of countably many thin pieces.
That one fact is Baire's theorem, and the proof is a single game of nested balls. Then it pays off four times in a row for linear maps between Banach spaces: Banach–Steinhaus (bounded at every point means uniformly bounded), the open mapping theorem, the bounded inverse theorem, and the closed graph theorem. All four fail the moment the space has a hole, and we'll see each one fail.
Here's the setup. A set $U$ is dense when it meets every nonempty open set. An open dense set can still have holes: $\mathbb{R} \setminus \{0\}$ is open and dense, and so is $\mathbb{R}$ minus the Cantor set. Intersect countably many of these and you might worry that the holes pile up until nothing is left. In a complete space they never do.
(Baire's Category Theorem.) Let $(X, d)$ be a complete metric space, and let $U_1, U_2, \ldots$ be open dense subsets of $X$. Then $\bigcap_{n=1}^{\infty} U_n$ is dense in $X$.
Proof. Write $B(x, r)$ for the open ball and $\overline{B(x, r)}$ for its closure. Let $W$ be a nonempty open set; we find a point of $W \cap \bigcap_n U_n$.
Since $U_1$ is dense, $U_1 \cap W$ is open and nonempty, so it contains some ball $B(x_1, s)$. Take $0 < r_1 < \min\{\frac{s}{2}, 1\}$. Then $\overline{B(x_1, r_1)} \subseteq \{y : d(x_1, y) \le r_1\} \subseteq B(x_1, s)$, so $\overline{B(x_1, r_1)} \subseteq U_1 \cap W$.
Suppose $x_{n-1}$ and $r_{n-1}$ are chosen. Since $U_n$ is dense, $U_n \cap B(x_{n-1}, r_{n-1})$ is open and nonempty, and the same argument gives $x_n$ and $0 < r_n < \frac{1}{n}$ with $$\overline{B(x_n, r_n)} \subseteq U_n \cap B(x_{n-1}, r_{n-1}).$$ So the closed balls are nested: $\overline{B(x_n, r_n)} \subseteq \overline{B(x_{n-1}, r_{n-1})} \subseteq \cdots \subseteq \overline{B(x_1, r_1)} \subseteq W$.
For $i, j \ge n$, both $x_i$ and $x_j$ lie in $B(x_n, r_n)$, so $d(x_i, x_j) < 2r_n < \frac{2}{n}$. The centers are Cauchy, and since $X$ is complete, $x_k \to x$ for some $x \in X$.
Fix $n$. Every $x_k$ with $k \ge n$ lies in the closed set $\overline{B(x_n, r_n)}$, so the limit does too: $x \in \overline{B(x_n, r_n)} \subseteq U_n \cap W$. This holds for every $n$, so $x \in W \cap \bigcap_n U_n$. $\blacksquare$
Two details carry the whole proof. Completeness hands over the limit $x$. The closures keep it inside: a limit of points of an open set can escape to the edge, but a limit of points of a closed ball can't. Here is the chase, with each $U_n$ the plane minus a closed set $F_n$ that is thin enough to dodge.
The ball sizes were chosen by hand there, and they had to shrink: $r_n < \frac{1}{n}$ is what forces the centers to be Cauchy. The proof also wrote $\overline{B(x, r)}$, not $\{y : d(x, y) \le r\}$, and that wasn't fussiness.
The closure of an open ball doesn't have to be the "closed ball" $\{y : d(x, y) \le r\}$. In the discrete metric ($d(x, y) = 1$ whenever $x \ne y$), $B(x, 1) = \{x\}$, which is already closed, while $\{y : d(x, y) \le 1\}$ is the whole space. The closure is always inside $\{d \le r\}$, and that containment is all the proof needs.
Now the hypothesis that does the real work. Run the same chase in a space with a hole.
Show that Baire's theorem fails in $\mathbb{Q}$ (with $d(x, y) = |x - y|$).
List the rationals: $\mathbb{Q} = \{q_1, q_2, q_3, \ldots\}$. Let $U_k = \mathbb{Q} \setminus \{q_k\}$.
Each $U_k$ is open in $\mathbb{Q}$, because the single point $\{q_k\}$ is closed. Each $U_k$ is dense in $\mathbb{Q}$: a nonempty open set of $\mathbb{Q}$ contains a set $(a, b) \cap \mathbb{Q}$, which has infinitely many points, so it has points other than $q_k$.
But $\bigcap_k U_k = \mathbb{Q} \setminus \{q_1, q_2, \ldots\} = \emptyset$, which is not dense. So the theorem fails, and the only hypothesis missing is completeness.
Where does the proof break? Exactly at "the centers converge". The nested balls can close in on a point that isn't there.
Baire says most of a complete space is generic. Like a mall.
The theorem is about open dense sets, but it upgrades for free to countable intersections of $G_\delta$ sets. That's the form we'll quote most.
In a complete metric space, if $G_1, G_2, \ldots$ are dense $G_\delta$ sets, then $\bigcap_n G_n$ is a dense $G_\delta$ set.
Proof. Write $G_n = \bigcap_k V_{n,k}$ with each $V_{n,k}$ open. Each $V_{n,k}$ contains the dense set $G_n$, so it is dense. Then $\bigcap_n G_n = \bigcap_{n, k} V_{n,k}$ is an intersection of countably many open dense sets: dense by Baire's theorem, and a $G_\delta$ by definition. $\blacksquare$
In particular, two dense $G_\delta$ sets in a complete space always meet. Compare $\mathbb{Q}$ and the irrationals: both dense, disjoint, and (this is the punchline of the corollary below) $\mathbb{Q}$ isn't a $G_\delta$. Now play the proof yourself. Each round drops a new closed nowhere dense set across your last ball, and your next closed ball has to dodge it.
Notice that you never ran out of room. A line, a circle, a grid of points: each one is closed and leaves whole open pieces of every ball free. That property has a name.
The sets we dodged were thin in a precise way: their closures contain no open set at all. Recall that the closure $\overline{E}$ is the smallest closed set containing $E$ (Closed Sets & Limit Points).
A subset $E$ of a metric space $X$ is nowhere dense if its closure $\overline{E}$ contains no nonempty open subset of $X$.
The word "nonempty" matters: every set contains the empty set, which is open. (The packet and the notes both dropped it.) Equivalently, $E$ is nowhere dense exactly when $X \setminus \overline{E}$ is open and dense. In $\mathbb{R}$, it means every interval contains a smaller interval that misses $\overline{E}$.
Which of these are nowhere dense in $\mathbb{R}$: (a) $\mathbb{Z}$; (b) $\{1/n : n \ge 1\}$; (c) the Cantor set; (d) $\mathbb{Q}$? And (e): is $\mathbb{Q} \times \{0\}$ nowhere dense in $\mathbb{R}^2$?
(a) Yes. $\mathbb{Z}$ is closed, and any interval $(a, b)$ contains an interval between two consecutive integers, or one with no integers at all, that misses $\mathbb{Z}$.
(b) Yes. The closure is $\{0\} \cup \{1/n\}$. An interval that contains $0$ also contains a gap $(\frac{1}{n+1}, \frac{1}{n})$, and an interval that doesn't contain $0$ meets only finitely many of the $1/n$, so it has a gap between them.
(c) Yes. The Cantor set is closed, and after $n$ steps it sits inside pieces of length $3^{-n}$. An interval of length $\ell$ can't fit inside a piece once $3^{-n} < \ell$, so it reaches into a removed middle third, which is an open interval missing the set.
(d) No. $\mathbb{Q}$ contains no interval, but its closure is all of $\mathbb{R}$.
(e) Yes. Its closure is the $x$-axis, and no disc fits inside a line. Being dense in a line doesn't make a set dense in the plane.
Part (d) is the trap: "contains no interval" and "nowhere dense" are different, because the definition is about the closure. Part (c) shows that nowhere dense sets can be uncountable. Hunt for gaps yourself. For a nowhere dense set there's always one, though sometimes you have to zoom in.
The fat Cantor set is the surprising one: it has measure $\frac{1}{2}$ and still has gaps everywhere. Being nowhere dense isn't about size in the measure sense. Your turn: find the nowhere dense set (or, when the question flips, the one that isn't).
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A single nowhere dense set is thin. Baire's theorem is really about countably many of them, so they get their own word.
A subset of a metric space $X$ is of the first category if it is a countable union of nowhere dense sets.
So $\mathbb{Q}$ is of the first category in $\mathbb{R}$, being a countable union of single points, even though it isn't nowhere dense. Everything else gets the other label.
A subset of a metric space $X$ is of the second category if it is not of the first category.
"Second category" just means "not first category." It's the participation trophy of topology.
First category sets can be dense, uncountable, even everywhere at once. A few to get used to:
Show that each of these is of the first category in $\mathbb{R}$: (a) $\mathbb{Q}$; (b) the algebraic numbers (roots of nonzero polynomials with integer coefficients); (c) $\bigcup_{q \in \mathbb{Q}} (C + q)$, where $C$ is the Cantor set.
(a) $\mathbb{Q} = \bigcup_n \{q_n\}$, and a single point is closed and holds no interval, so it is nowhere dense.
(b) There are countably many nonzero integer polynomials (countably many for each degree and each bound on the coefficients), and each has finitely many roots. So the algebraic numbers are countable, a countable union of points.
(c) Translation $x \mapsto x + q$ is a homeomorphism of $\mathbb{R}$, so it carries closed sets to closed sets and intervals to intervals. Each $C + q$ is therefore closed with no interval inside: nowhere dense. There are countably many of them. This set is uncountable and dense, and it is still of the first category.
Read Baire's theorem through these words and it says a complete space is never thin.
A nonempty complete metric space $X$ is of the second category (in itself). More generally, the complement of a set of the first category in $X$ is dense.
Proof. Let $E = \bigcup_n E_n$ with each $E_n$ nowhere dense. The set $X \setminus \overline{E_n}$ is open, and it is dense: a nonempty open set $W$ that missed it would satisfy $W \subseteq \overline{E_n}$, which the definition forbids. By Baire's theorem, $D = \bigcap_n (X \setminus \overline{E_n})$ is dense. A point of $D$ lies in no $\overline{E_n}$, hence in no $E_n$, so $D \subseteq X \setminus E$, and $X \setminus E$ is dense.
If $E$ were all of $X$, then $X \setminus E = \emptyset$ would be dense in the nonempty space $X$, which is impossible. So $X$ is not of the first category. $\blacksquare$
Here it is at work in the plane.
Show that $\mathbb{R}^2$ is not a countable union of straight lines.
A line $\ell$ is closed. It holds no disc: a disc contains three points that aren't collinear, and a line doesn't. So each line is nowhere dense, and a countable union of lines is of the first category. But $\mathbb{R}^2$ is complete and nonempty, so by the corollary it is of the second category. So no countable family of lines covers it.
The same argument works for circles, graphs of continuous functions, or any countable mix of closed sets with empty interior. The second sentence of the corollary is the one to use. It says a first category set can't contain an open ball (its complement would miss the ball), so in $\mathbb{R}$ every set containing an interval is of the second category. And a set and its complement can't both be first category, since their union would be. So the irrationals are of the second category: their complement $\mathbb{Q}$ is first category. Baire's theorem also runs backwards, and that's the form two of the big theorems below use.
Let $X$ be a complete metric space and $U_1, U_2, \ldots$ open subsets of $X$. If $\bigcap_n U_n$ is not dense, then some $U_n$ is not dense, so $X \setminus \overline{U_n}$ is a nonempty open set (it contains a ball).
Proof. If every $U_n$ were dense, Baire's theorem would make $\bigcap_n U_n$ dense. So some $U_n$ is not dense, which means some nonempty open set misses $U_n$, and hence misses $\overline{U_n}$ too (an open set that misses a set misses its closure). That open set lies in $X \setminus \overline{U_n}$, so this set is open and nonempty. $\blacksquare$
Now a warning. "First category" and "measure zero" both feel like "small", and they have nothing to do with each other.
Find a dense $G_\delta$ subset of $\mathbb{R}$ with Lebesgue measure $0$. Conclude that $\mathbb{R}$ is the union of a set of the first category and a set of measure $0$.
List the rationals $q_1, q_2, \ldots$ and let $$U_n = \bigcup_{k=1}^{\infty} \left(q_k - 2^{-k-n}, \ q_k + 2^{-k-n}\right), \qquad G = \bigcap_{n=1}^{\infty} U_n.$$ Each $U_n$ is open and contains $\mathbb{Q}$, so it is dense. So $G$ is a dense $G_\delta$ (it is a $G_\delta$ by definition, and dense by Baire's theorem).
Its measure: $m(G) \le m(U_n) \le \sum_k 2 \cdot 2^{-k-n} = 2^{1-n}$ for every $n$, so $m(G) = 0$.
Its complement: $\mathbb{R} \setminus G = \bigcup_n (\mathbb{R} \setminus U_n)$, and each $\mathbb{R} \setminus U_n$ is closed and contains no interval (an interval would have to miss the dense set $U_n$). So $\mathbb{R} \setminus G$ is of the first category, and $\mathbb{R} = (\mathbb{R} \setminus G) \cup G$.
So the first category half carries all the measure, and the measure zero half is of the second category. It goes the other way too: the fat Cantor set is nowhere dense and has measure $\frac{1}{2}$. Here are both at once.
Measure asks how much room a set takes up. Category asks whether you can dodge it. A set can be huge by one measure and tiny by the other.
Other books use other words. A first category set is often called meager, a second category set nonmeager, and a set whose complement is first category is called residual or comeager. A space in which every countable intersection of open dense sets is dense is called a Baire space; Baire's theorem says complete metric spaces are Baire spaces. This page sticks to first and second category.
Here is the corollary that the Riesz page proved by hand for countable sets dense in an interval.
The Proposition on the Riesz page showed that a countable set dense in an interval is not a $G_\delta$, with a nested-interval argument. That argument was Baire's proof in disguise.
Let $X$ be a nonempty complete metric space with no isolated points. Then no countable dense subset of $X$ is a $G_\delta$.
Proof. Suppose $E = \{x_1, x_2, \ldots\}$ is countable and dense (if $E$ is finite, repeat its last point forever), and $E = \bigcap_k V_k$ with each $V_k$ open. Each $V_k$ contains $E$, so it is dense.
A single point $\{x\}$ is closed, and it is nowhere dense: a nonempty open subset of $\{x\}$ would be $\{x\}$ itself, which would make $x$ an isolated point. So $X \setminus \{x\}$ is open and dense. Let $$U_k = V_k \setminus \{x_1, \ldots, x_k\} = V_k \cap (X \setminus \{x_1\}) \cap \cdots \cap (X \setminus \{x_k\}).$$ A finite intersection of open dense sets is open and dense (if $A$ and $B$ are open and dense and $W$ is open and nonempty, then $W \cap A$ is open and nonempty, so it meets $B$). So each $U_k$ is open and dense.
But every point of $\bigcap_k U_k$ lies in $\bigcap_k V_k = E$, and each $x_n$ was removed from $U_n$. So $\bigcap_k U_k = \emptyset$, which is not dense in the nonempty space $X$. That contradicts Baire's theorem. $\blacksquare$
So $\mathbb{Q}$ is not a $G_\delta$ in $\mathbb{R}$, and neither is any countable dense set, in $\mathbb{R}^n$ or the Cantor set or any complete space without isolated points. The same trick of removing points one at a time proves something you've known since calculus, with a brand new proof.
Prove that a nonempty complete metric space with no isolated points is uncountable. Conclude that $\mathbb{R}$ and the Cantor set are uncountable.
Proof. Suppose $X = \{x_1, x_2, \ldots\}$ is countable. ($X$ can't be finite: in a finite metric space every point is isolated, since the ball of radius less than the smallest distance is a single point.) Each $U_n = X \setminus \{x_n\}$ is open, because points are closed. It is dense: a nonempty open set $W$ missing $U_n$ would satisfy $W \subseteq \{x_n\}$, so $W = \{x_n\}$ would be open, and $x_n$ would be isolated. By Baire's theorem $\bigcap_n U_n$ is dense in the nonempty space $X$, so it is nonempty. But $\bigcap_n U_n = X \setminus \{x_1, x_2, \ldots\} = \emptyset$. This contradiction shows $X$ is uncountable.
$\mathbb{R}$ is complete with no isolated points. The Cantor set is closed in $\mathbb{R}$, so it is complete. It has no isolated points: given $x$ in it and $\varepsilon > 0$, pick $n$ with $3^{-n} < \varepsilon$. Then $x$ lies in one of the stage-$n$ pieces, an interval of length $3^{-n}$ whose two endpoints stay in the Cantor set forever. At least one endpoint differs from $x$, and it is within $\varepsilon$ of $x$. Both are uncountable. $\blacksquare$
Now sort sets into all four bins at once. In $\mathbb{R}$ every set lands in exactly one: nowhere dense; first category but not nowhere dense; second category with no interval inside; or containing an interval.
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Now the payoff for linear maps. Take a whole family of bounded linear maps $\Lambda_\alpha\colon X \to Y$. Each one is bounded, but maybe their norms aren't bounded together. Banach–Steinhaus says there are only two possibilities: either one bound $M$ works for all of them, or they blow up at most points of $X$.
(Banach–Steinhaus.) Let $X$ be a Banach space, $Y$ a normed space, and $\{\Lambda_\alpha : \alpha \in A\}$ a collection of bounded linear maps $X \to Y$. Then exactly one of these holds:
Proof. Let $\varphi_\alpha(x) = \|\Lambda_\alpha x\|$ and $\varphi(x) = \sup_\alpha \varphi_\alpha(x) \in [0, \infty]$. Each $\varphi_\alpha$ is continuous (a composition of continuous maps), so for each $n \in \mathbb{N}$, $$V_n = \{x : \varphi(x) > n\} = \bigcup_{\alpha \in A} \{x : \varphi_\alpha(x) > n\}$$ is a union of open sets, hence open. (In other words, $\varphi$ is lower semicontinuous.)
Case 1: every $V_n$ is dense. By Baire's theorem, $S = \bigcap_n V_n$ is a dense $G_\delta$, and $\varphi(x) > n$ for every $n$ when $x \in S$. That is (b).
Case 2: some $V_N$ is not dense. Then some nonempty open set misses $V_N$, so there are $y \in X$ and $\delta > 0$ with $B(y, \delta) \subseteq X \setminus V_N$. On that ball, $\|\Lambda_\alpha x\| \le N$ for every $\alpha$. If $\|z\| < \delta$, then $y + z$ and $y$ are both in the ball, so by linearity $$\|\Lambda_\alpha z\| = \|\Lambda_\alpha(y + z) - \Lambda_\alpha y\| \le \|\Lambda_\alpha(y + z)\| + \|\Lambda_\alpha y\| \le 2N.$$ Now let $\|z\| \le 1$ and $0 < t < \delta$. Then $\|tz\| < \delta$, so $\|\Lambda_\alpha z\| = \frac{1}{t}\|\Lambda_\alpha(tz)\| \le \frac{2N}{t}$. Letting $t \to \delta$ gives $\|\Lambda_\alpha z\| \le \frac{2N}{\delta}$. So $\|\Lambda_\alpha\| \le \frac{2N}{\delta} = M$ for every $\alpha$. That is (a).
Not both. If (a) holds, then $\|\Lambda_\alpha x\| \le M\|x\| < \infty$ for every $x$, so the set in (b) is empty, and an empty set is not dense in $X \ne \emptyset$. $\blacksquare$
The packet's statement says only "a $G_\delta$ set $S$" in (b). Without "dense" the theorem says nothing, since $S = \emptyset$ is a $G_\delta$. Geometrically: either one ball in $Y$ holds the image of the unit ball under every $\Lambda_\alpha$, or the images of most single points escape every ball. Case 2 of the proof uses the same two moves as bounded ⇔ continuous.
On Banach Spaces & Bounded Operators, continuity at one point became a bound on the unit ball: slide the small ball to the origin, then scale it up. Banach–Steinhaus does the same thing to a whole family at once, and Baire's theorem is what finds the small ball.
The bound $2N/\delta$ is usually wasteful (there, the real norms are under $0.75$ and the bound is $4$), but the proof never needs a good bound, just a finite one. Now watch alternative (b) happen.
On $c_0$ (with $\|x\|_\infty$), let $\Lambda_N x = x_1 + \cdots + x_N$. Which alternative of Banach–Steinhaus holds? Find an explicit $x \in c_0$ with $\sup_N |\Lambda_N x| = \infty$.
$|\Lambda_N x| \le N\|x\|_\infty$, and $x = (1, \ldots, 1, 0, 0, \ldots)$ with $N$ ones has norm $1$ and $\Lambda_N x = N$. So $\|\Lambda_N\| = N$, which is unbounded, and (a) fails. So (b) holds: the $x$ with unbounded partial sums form a dense $G_\delta$ in $c_0$.
An explicit one: build $x$ from blocks, where block $j$ is $j$ entries equal to $\frac{1}{j}$. The entries tend to $0$, so $x \in c_0$. Each block adds exactly $j \cdot \frac{1}{j} = 1$ to the partial sums, so at the end of block $j$, $\Lambda_N x = j$, which is unbounded.
The theorem guarantees such vectors exist without building any. Here's the block vector anyway, entries shrinking and partial sums climbing.
Turn the theorem around and you get its most quoted form: bounded at every point forces a uniform bound.
(Uniform Boundedness Principle.) Let $X$ be a Banach space, $Y$ a normed space, and $\{\Lambda_\alpha\}$ bounded linear maps $X \to Y$. If $\sup_\alpha \|\Lambda_\alpha x\| < \infty$ for every $x \in X$, then $\sup_\alpha \|\Lambda_\alpha\| < \infty$.
Proof. The hypothesis says the set in (b) of Banach–Steinhaus is empty, so (b) fails. So (a) holds. $\blacksquare$
Completeness of $X$ is not decoration here.
On $c_{00}$ (finitely supported sequences, with the sup norm), let $\Lambda_n x = n x_n$. Every $x$ is eventually $0$, so $\sup_n |\Lambda_n x|$ is a max over finitely many terms: finite. But $\|\Lambda_n\| = n$ (test on $e_n$). Bounded at every point, not uniformly bounded. $c_{00}$ is not complete, and the principle fails.
Try it. Pick a space, a family, and a vector, and compare the norms $\|\Lambda_n\|$ with the values $|\Lambda_n x|$.
On $c_0$ most of the vectors you can name don't blow up. That doesn't contradict "most points of $X$ blow up": the vectors we can write down are special. The set $S$ in (b) is dense, so a blow-up vector sits within $\varepsilon$ of every vector, and you can see why.
Adding $\varepsilon/\sqrt{k}$ costs only $\varepsilon$ in the sup norm, but its partial sums grow like $2\varepsilon\sqrt{N}$, so they always win eventually. The principle earns its keep in proofs like the next one, where you can't compute anything directly.
On Banach Spaces & Bounded Operators, the functional $\xi \mapsto \sum_{i \le N} \alpha_i \xi_i$ on $c_0$ was shown to have norm exactly $\sum_{i \le N} |\alpha_i|$, by testing on the signs of the $\alpha_i$.
Suppose $\sum_i \alpha_i \xi_i$ converges for every sequence $(\xi_i)$ with $\xi_i \to 0$. Prove that $\sum_i |\alpha_i| < \infty$.
Proof. On the Banach space $c_0$, let $\Lambda_N \xi = \sum_{i \le N} \alpha_i \xi_i$. Each $\Lambda_N$ is a bounded linear functional, with $\|\Lambda_N\| = \sum_{i \le N} |\alpha_i|$ (the bound $|\Lambda_N \xi| \le \sum_{i \le N} |\alpha_i| \, \|\xi\|_\infty$, and equality at $\xi_i = \overline{\alpha_i}/|\alpha_i|$ for $i \le N$ with $\alpha_i \ne 0$, and $\xi_i = 0$ otherwise).
For each $\xi \in c_0$, the sequence $(\Lambda_N \xi)_N$ is the sequence of partial sums of a convergent series, so it is bounded: $\sup_N |\Lambda_N \xi| < \infty$. By the uniform boundedness principle, $\sup_N \sum_{i \le N} |\alpha_i| = \sup_N \|\Lambda_N\| < \infty$. The partial sums of $\sum_i |\alpha_i|$ are bounded, so it converges. $\blacksquare$
Notice what the proof never did: construct a bad $\xi$. If $\sum |\alpha_i| = \infty$, a sequence $\xi \to 0$ with $\sum \alpha_i \xi_i$ divergent exists, but the principle hands it over without building it. The same pattern works in $\ell^2$.
Let $(a_n)$ be positive numbers such that $\sum a_n b_n < \infty$ whenever $b_n \ge 0$ and $\sum b_n^2 < \infty$. Prove that $\sum a_n^2 < \infty$.
Proof. On the Hilbert space $\ell^2$, which is complete, let $\Lambda_N b = \sum_{n \le N} a_n b_n$. By Cauchy–Schwarz, $|\Lambda_N b| \le \left(\sum_{n \le N} a_n^2\right)^{1/2} \|b\|_2$, with equality at $b = (a_1, \ldots, a_N, 0, \ldots)$. So $\|\Lambda_N\| = \left(\sum_{n \le N} a_n^2\right)^{1/2}$.
For any $b \in \ell^2$, the sequence $(|b_n|)$ is nonnegative and square-summable, so by hypothesis $\sum_n a_n |b_n| < \infty$, and $|\Lambda_N b| \le \sum_n a_n |b_n|$ for every $N$. So $\sup_N |\Lambda_N b| < \infty$ for every $b$, and the uniform boundedness principle gives $\sup_N \|\Lambda_N\| = M < \infty$. Then $\sum_{n \le N} a_n^2 \le M^2$ for every $N$, so $\sum a_n^2 \le M^2 < \infty$. $\blacksquare$
Banach–Steinhaus gives a uniform bound. Once you have one, convergence on a dense set spreads to the whole space, with no completeness needed in the domain at all.
Let $X$ be a normed space, $Y$ a Banach space, and $\Lambda_n\colon X \to Y$ bounded linear maps with $\|\Lambda_n\| \le M < \infty$ for all $n$. Suppose there is a dense set $E \subseteq X$ such that $(\Lambda_n x)$ converges for each $x \in E$. Prove that $(\Lambda_n x)$ converges for each $x \in X$.
Proof. If $M = 0$, every $\Lambda_n$ is $0$. Otherwise fix $x \in X$ and $\varepsilon > 0$. Since $E$ is dense, pick $e \in E$ with $\|x - e\| < \frac{\varepsilon}{3M}$. Since $(\Lambda_n e)$ converges, it is Cauchy: there is $N$ with $\|\Lambda_n e - \Lambda_m e\| < \frac{\varepsilon}{3}$ for $n, m \ge N$. Then for $n, m \ge N$, $$\|\Lambda_n x - \Lambda_m x\| \le \|\Lambda_n(x - e)\| + \|\Lambda_n e - \Lambda_m e\| + \|\Lambda_m(e - x)\| < M\frac{\varepsilon}{3M} + \frac{\varepsilon}{3} + M\frac{\varepsilon}{3M} = \varepsilon.$$ So $(\Lambda_n x)$ is Cauchy in the Banach space $Y$, and it converges. $\blacksquare$
Put the two together, with $X$ a Banach space: if $(\Lambda_n x)$ converges for every $x$, the limits are bounded, so the norms are bounded by the principle; and if the norms are bounded, convergence on a dense set is enough.
On Fourier Series, Banach–Steinhaus answers a question that stumped people for decades: does the Fourier series of every continuous function converge at $0$? The partial sums at $0$ are functionals on $C(T)$ whose norms grow like $\log n$, so no: they blow up on a dense $G_\delta$ of continuous functions.
One family of functionals gets a name of its own, because the Spicy Bonus at the bottom of the page lives there: the bounded linear functionals on a normed space $X$ form its dual space $X^*$, normed by $\|\Lambda\| = \sup\{|\Lambda x| : \|x\| \le 1\}$. (Its official home is Hahn–Banach & the Poisson Integral.) By the Bounded Operators page, the dual of $c_0$ is $\ell^1$, and the dual of $\ell^1$ is $\ell^\infty$.
For now, practice the dichotomy: compute the norms, take the sup, and read off which alternative holds.
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A map is an open map when it sends open sets to open sets. In finite dimensions, a linear map onto $\mathbb{R}^n$ is invertible after cutting down its kernel, and openness is easy. In Banach spaces, "onto" still forces "open", but proving it takes Baire's theorem and a geometric series.
(Open Mapping Theorem.) Let $X$ and $Y$ be Banach spaces. If $\Lambda\colon X \to Y$ is a bounded linear map onto $Y$, then $\Lambda$ is an open map.
Proof. Write $B_X(x, r)$ and $B_Y(y, r)$ for open balls. The heart of the proof is
$(\ast)$ there is $\delta > 0$ with $B_Y(0, \delta) \subseteq \Lambda(B_X(0, 1))$.
$(\ast)$ implies open. Let $U \subseteq X$ be open and $y \in \Lambda(U)$, say $y = \Lambda u$ with $u \in U$, and pick $\varepsilon > 0$ with $B_X(u, \varepsilon) \subseteq U$. If $\|w - y\| < \varepsilon\delta$, then $\frac{1}{\varepsilon}(w - y) \in B_Y(0, \delta)$, so by $(\ast)$ there is $x \in B_X(0, 1)$ with $\Lambda x = \frac{1}{\varepsilon}(w - y)$. Then $v = u + \varepsilon x$ has $\|v - u\| < \varepsilon$, so $v \in U$, and $\Lambda v = y + (w - y) = w$. So $B_Y(y, \varepsilon\delta) \subseteq \Lambda(U)$, and $\Lambda(U)$ is open.
Step 1: a ball in a closure. Since $\Lambda$ is onto, $Y = \bigcup_n \Lambda(B_X(0, n))$. The space $Y$ is complete, so it is of the second category, and some $\Lambda(B_X(0, n))$ is not nowhere dense: its closure contains a ball. The map $y \mapsto y/n$ is a homeomorphism of $Y$ that carries $\Lambda(B_X(0, n))$ onto $\Lambda(B_X(0, 1))$ (by linearity), and so carries closure onto closure. So $\overline{\Lambda(B_X(0, 1))} \supset B_Y(y_0, 4r)$ for some $y_0 \in Y$ and $r > 0$.
Step 2: centering it. Since $y_0$ lies in that closure, the ball $B_Y(y_0, 2r)$ meets $\Lambda(B_X(0, 1))$: pick $y' = \Lambda x'$ in it, with $\|x'\| < 1$. Then $B_Y(y', 2r) \subseteq B_Y(y_0, 4r) \subseteq \overline{\Lambda(B_X(0, 1))}$. For $\|y\| < 2r$, $y + y'$ lies in that closure, so $$y = (y + y') - \Lambda x' \in \overline{\Lambda(B_X(0, 1))} - \Lambda x' = \overline{\Lambda(B_X(0, 1) - x')} \subseteq \overline{\Lambda(B_X(0, 2))},$$ using that translation is a homeomorphism and $B_X(0, 1) - x' \subseteq B_X(0, 2)$. Scaling by $\frac{1}{2}$, and then by any $t > 0$: $$B_Y(0, tr) \subseteq \overline{\Lambda(B_X(0, t))} \quad \text{for every } t > 0.$$
Step 3: correcting the aim. Let $\delta = \frac{r}{2}$ and $y \in B_Y(0, \delta)$. With $t = \frac{1}{2}$, $y \in \overline{\Lambda(B_X(0, \frac{1}{2}))}$, so there is $x_1 \in B_X(0, \frac{1}{2})$ with $\|y - \Lambda x_1\| < \frac{r}{4}$. Then $y - \Lambda x_1 \in B_Y(0, \frac{r}{4}) \subseteq \overline{\Lambda(B_X(0, \frac{1}{4}))}$, so there is $x_2 \in B_X(0, \frac{1}{4})$ with $\|y - \Lambda x_1 - \Lambda x_2\| < \frac{r}{8}$. Inductively, there are $x_n \in B_X(0, 2^{-n})$ with $$\left\|y - \Lambda(x_1 + \cdots + x_n)\right\| < r \, 2^{-n-1}.$$
Step 4: summing. $\sum_n \|x_n\| < \sum_n 2^{-n} = 1$, so the series converges absolutely, and since $X$ is complete it converges to some $x$, with $\|x\| \le \|x_1\| + \sum_{n \ge 2} \|x_n\| < \frac{1}{2} + \frac{1}{2} = 1$. Since $\Lambda$ is continuous, $\Lambda(x_1 + \cdots + x_n) \to \Lambda x$, while by Step 3 it also tends to $y$. So $y = \Lambda x \in \Lambda(B_X(0, 1))$, which proves $(\ast)$. $\blacksquare$
The proof has two halves, and it helps to see them separately. The first half says $(\ast)$ is all that "open" really means: if the image of the unit ball holds a ball around $0$, then translating and scaling handle every other open set. Losing "onto" breaks exactly that.
Surjective, bounded, between two Banach spaces, and it still needs four steps and a geometric series to admit it's open.
The second half is where the work is. Baire only says the closure of $\Lambda(B_X(0, 1))$ holds a ball, so we can aim at $y$ but never hit it exactly. Each miss is half the last, and completeness of $X$ adds up the corrections.
In $\mathbb{R}^2$ you can see the whole story at once: the largest disc inside $\Lambda(\text{unit disc})$ has radius $\delta$, the smallest singular value of $\Lambda$, and onto means exactly $\delta > 0$.
The readout $\|\Lambda^{-1}\| = 1/\delta$ is a preview of the next corollary. Before that, play Step 3 by hand. You only get grid points, the way the proof only gets points of a dense set.
Every round looked the same after the zoom, and that self-similarity is the geometric series. Unpacked, $(\ast)$ says something concrete about preimages: you can always pick one that isn't much longer than its image.
Let $\Lambda\colon X \to Y$ be a bounded linear map of a Banach space onto a Banach space. Show that there is a constant $C$ such that every $y \in Y$ equals $\Lambda x$ for some $x$ with $\|x\| \le C\|y\|$.
By $(\ast)$ in the proof of the open mapping theorem, $B_Y(0, \delta) \subseteq \Lambda(B_X(0, 1))$ for some $\delta > 0$. For $y = 0$, take $x = 0$. For $y \ne 0$, the vector $y' = \frac{\delta}{2\|y\|} y$ has $\|y'\| = \frac{\delta}{2} < \delta$, so $y' = \Lambda x'$ with $\|x'\| < 1$. Then $x = \frac{2\|y\|}{\delta} x'$ has $\Lambda x = y$ and $\|x\| < \frac{2}{\delta} \|y\|$. So $C = \frac{2}{\delta}$ works.
When $\Lambda$ is also one-to-one, that preimage is the only one, and the constant becomes a bound on $\Lambda^{-1}$. That's the most-used consequence of the theorem, and its proof is one line long.
(Bounded Inverse Theorem.) Let $X$ and $Y$ be Banach spaces and $\Lambda\colon X \to Y$ a bounded linear bijection. Then $\Lambda^{-1}$ is bounded: there is $c > 0$ with $\|\Lambda x\| \ge c\|x\|$ for every $x$.
Proof. $\Lambda^{-1}$ is linear: if $\Lambda x = y$ and $\Lambda x' = y'$, then $\Lambda(\alpha x + x') = \alpha y + y'$. For an open $U \subseteq X$, $(\Lambda^{-1})^{-1}(U) = \Lambda(U)$ is open by the open mapping theorem, so $\Lambda^{-1}$ is continuous, hence bounded. Then $\|x\| = \|\Lambda^{-1} \Lambda x\| \le \|\Lambda^{-1}\| \, \|\Lambda x\|$, so $c = 1/\|\Lambda^{-1}\|$ works. $\blacksquare$
The notes state this for any bijection, and "bounded linear" has to be there: the open mapping theorem needs it. Completeness on both sides has to be there too.
Show that the identity map $I\colon (\ell^1, \|\cdot\|_1) \to (\ell^1, \|\cdot\|_\infty)$ is a bounded linear bijection whose inverse is unbounded. Which hypothesis of the bounded inverse theorem fails?
$\|x\|_\infty \le \sum_n |x_n| = \|x\|_1$, so $\|I\| \le 1$, and $I$ is a linear bijection. For the inverse, $x = e_1 + \cdots + e_n$ has $\|x\|_\infty = 1$ and $\|x\|_1 = n$, so $\|I^{-1}\| \ge n$ for every $n$.
The domain is a Banach space, so the target must be the problem: $(\ell^1, \|\cdot\|_\infty)$ is not complete. Indeed, the partial sums of $\sum_n \frac{1}{n} e_n$ are Cauchy in the sup norm (their differences have sup norm at most $\frac{1}{m}$ past $m$), but any limit would have to equal $(\frac{1}{n})$ in every coordinate, and $(\frac{1}{n}) \notin \ell^1$.
So a one-sided bound between two norms can't be reversed in general. Between two complete norms, it always can.
Let $\|\cdot\|_a$ and $\|\cdot\|_b$ be two norms on a vector space $X$, each making $X$ a Banach space, with $\|x\|_a \le C\|x\|_b$ for all $x$. Prove that there is $C'$ with $\|x\|_b \le C'\|x\|_a$ for all $x$, so the norms are equivalent.
Proof. The identity $I\colon (X, \|\cdot\|_b) \to (X, \|\cdot\|_a)$ is linear and bijective, and $\|Ix\|_a \le C\|x\|_b$ says it is bounded. Both spaces are Banach, so by the bounded inverse theorem $I^{-1}$ is bounded: $\|x\|_b = \|I^{-1}x\|_b \le \|I^{-1}\| \, \|x\|_a$. Take $C' = \|I^{-1}\|$. $\blacksquare$
The last consequence gives a new way to prove a linear map is bounded, and it's often the easiest one. Give $X \times Y$ the norm $\|(x, y)\| = \|x\| + \|y\|$. A sequence converges in this norm exactly when both coordinates converge, and a Cauchy sequence has Cauchy coordinates, so $X \times Y$ is a Banach space when $X$ and $Y$ are.
The graph of a linear map $\Lambda\colon X \to Y$ is $\mathcal{G}(\Lambda) = \{(x, \Lambda x) : x \in X\} \subseteq X \times Y$.
Since $\Lambda$ is linear, $\mathcal{G}(\Lambda)$ is a subspace of $X \times Y$. Continuity always makes it closed. For linear maps between Banach spaces, the converse holds too.
(Closed Graph Theorem.) Let $X$ and $Y$ be Banach spaces and $\Lambda\colon X \to Y$ linear. Then $\Lambda$ is bounded if and only if $\mathcal{G}(\Lambda)$ is closed in $X \times Y$.
Proof. ($\implies$) Let $(x_n, \Lambda x_n) \to (x, y)$ in $X \times Y$. Then $x_n \to x$ and $\Lambda x_n \to y$. A bounded map is continuous, so $\Lambda x_n \to \Lambda x$, and limits are unique, so $y = \Lambda x$. So $(x, y) \in \mathcal{G}(\Lambda)$, and the graph is closed.
($\impliedby$) $\mathcal{G}(\Lambda)$ is a closed subspace of the Banach space $X \times Y$, so it is a Banach space. Let $\pi\colon \mathcal{G}(\Lambda) \to X$, $\pi(x, \Lambda x) = x$. It is linear and bijective, and bounded since $\|x\| \le \|x\| + \|\Lambda x\| = \|(x, \Lambda x)\|$. By the bounded inverse theorem, $\pi^{-1}\colon x \mapsto (x, \Lambda x)$ is bounded: there is $C$ with $\|x\| + \|\Lambda x\| \le C\|x\|$ for all $x$. In particular $\|\Lambda x\| \le C\|x\|$, so $\Lambda$ is bounded. $\blacksquare$
Spell out what "closed graph" asks you to check, and you'll see why the theorem is so useful.
To prove $\Lambda$ is continuous directly, you'd take $x_n \to x$ and show $\Lambda x_n \to \Lambda x$, including that $\Lambda x_n$ converges at all. The closed graph theorem lets you assume $\Lambda x_n$ converges, say to $y$, and asks only that $y = \Lambda x$. That extra hypothesis is free, and it's usually the whole difficulty.
Here's the classic use: a symmetry condition that never mentions norms forces boundedness.
Let $H$ be a Hilbert space and $\Lambda\colon H \to H$ linear with $\langle \Lambda x, y \rangle = \langle x, \Lambda y \rangle$ for all $x, y \in H$. Show that $\Lambda$ is bounded.
Hilbert spaces are Banach spaces, so the closed graph theorem applies. Suppose $x_n \to x$ and $\Lambda x_n \to z$. For every $y \in H$, by continuity of the inner product, $$\langle z, y \rangle = \lim_n \langle \Lambda x_n, y \rangle = \lim_n \langle x_n, \Lambda y \rangle = \langle x, \Lambda y \rangle = \langle \Lambda x, y \rangle.$$ So $\langle z - \Lambda x, y \rangle = 0$ for every $y$; taking $y = z - \Lambda x$ gives $z = \Lambda x$. The graph is closed, so $\Lambda$ is bounded.
The proof never estimated $\|\Lambda x\|$ at all: it used the free hypothesis that $\Lambda x_n$ converges. (This is the Hellinger–Toeplitz theorem.) Linearity matters here. For a function $f\colon \mathbb{R} \to \mathbb{R}$, a closed graph doesn't even give continuity: take $f(x) = 1/x$ for $x \ne 0$ and $f(0) = 0$. Points of the graph near the $y$-axis run off to infinity, so they never converge to a missing point.
$1/x$ with $f(0) = 0$: a closed graph and a function that answers "I'm fine" while running off to infinity.
Sort some functions from $\mathbb{R}$ to $\mathbb{R}$ by what their graphs do. One of the four bins is a trick.
The empty bin is the easy direction of the theorem: continuous maps always have closed graphs. The prelim's version of the hard direction just restates "closed" with sequences. (The packet's version uses $y$ without saying what it is; here it is fixed.)
Let $X$ and $Y$ be Banach spaces and $\Lambda\colon X \to Y$ linear, with this property: whenever $x_n \to x$ in $X$ and $\Lambda x_n \to y$ in $Y$, then $y = \Lambda x$. Prove that $\Lambda$ is continuous.
Proof. We show $\mathcal{G}(\Lambda)$ is closed. Let $(x_n, \Lambda x_n) \to (x, y)$ in $X \times Y$. Then $\|x_n - x\| + \|\Lambda x_n - y\| \to 0$, so $x_n \to x$ and $\Lambda x_n \to y$. By the property, $y = \Lambda x$, so $(x, y) \in \mathcal{G}(\Lambda)$. By the closed graph theorem $\Lambda$ is bounded, and bounded linear maps are continuous. $\blacksquare$
That's all four theorems. Each one needs completeness somewhere, and each one has a hypothesis you can't drop. Check that you can tell them apart.
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Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: claims about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.
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Next, the proofs. Each one is a few lines from a proof on this page with one step missing.
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Then the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the types that need their answer choices will say so.
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Last, the Spicy Bonus: the dual space is always complete, and a sequence that every functional sees as bounded really is bounded. Part (b) needs one fact from a later page.
On Hahn–Banach & the Poisson Integral, a corollary of the Hahn–Banach theorem says that for every $x \ne 0$ in a normed space $X$ there is $\Lambda \in X^*$ with $\|\Lambda\| = 1$ and $\Lambda(x) = \|x\|$. The Spicy Bonus quotes that fact and proves everything else.
Let $X$ be a normed space, and give $X^*$ the norm $\|\Lambda\| = \sup\{|\Lambda(x)| : \|x\| \le 1\}$.
Proof. (a) $X^*$ is a vector space, and $\|\Lambda\|$ is the operator norm, so it is a norm. Let $(\Lambda_n)$ be Cauchy in $X^*$. For each $x$, $|\Lambda_n(x) - \Lambda_m(x)| \le \|\Lambda_n - \Lambda_m\| \, \|x\|$, so $(\Lambda_n(x))$ is Cauchy in $\mathbb{C}$ and converges; call the limit $\Lambda(x)$. A pointwise limit of linear maps is linear, so $\Lambda$ is linear. A Cauchy sequence is bounded, say $\|\Lambda_n\| \le K$, so $|\Lambda(x)| = \lim_n |\Lambda_n(x)| \le K\|x\|$, and $\Lambda \in X^*$.
It remains to show $\|\Lambda_n - \Lambda\| \to 0$. Given $\varepsilon > 0$, pick $N$ with $\|\Lambda_n - \Lambda_m\| < \varepsilon$ for $n, m \ge N$. For $\|x\| \le 1$ and $n, m \ge N$, $|\Lambda_n(x) - \Lambda_m(x)| < \varepsilon$; let $m \to \infty$ to get $|\Lambda_n(x) - \Lambda(x)| \le \varepsilon$. Taking the sup over $x$, $\|\Lambda_n - \Lambda\| \le \varepsilon$ for $n \ge N$. So $X^*$ is complete. Only the completeness of $\mathbb{C}$ was used.
(b) $\lambda_x(\alpha \Lambda + \Lambda') = \alpha \Lambda(x) + \Lambda'(x)$, so $\lambda_x$ is linear, and $|\lambda_x(\Lambda)| = |\Lambda(x)| \le \|\Lambda\| \, \|x\|$ gives $\|\lambda_x\| \le \|x\|$. If $x = 0$, equality holds. If $x \ne 0$, the Hahn–Banach fact gives $\Lambda \in X^*$ with $\|\Lambda\| = 1$ and $\Lambda(x) = \|x\|$, so $\|\lambda_x\| \ge |\lambda_x(\Lambda)| = \|x\|$. So $\|\lambda_x\| = \|x\|$.
(c) By (a), $X^*$ is a Banach space, and by (b) each $\lambda_{x_n}$ is a bounded linear functional on it. The hypothesis says $\sup_n |\lambda_{x_n}(\Lambda)| = \sup_n |\Lambda(x_n)| < \infty$ for every $\Lambda \in X^*$. By the uniform boundedness principle, applied on the Banach space $X^*$, $\sup_n \|\lambda_{x_n}\| < \infty$. By (b), $\|\lambda_{x_n}\| = \|x_n\|$, so $(\|x_n\|)$ is bounded. Notice that $X$ itself never had to be complete: the principle ran on $X^*$, which always is. $\blacksquare$