Topology was the first preliminary exam I took as a graduate student. I hope you enjoy this subject as much as I do!
Here's the big idea. In calculus, "open interval," "close to," and "continuous" were all defined with distances: $|x - a| < \delta$ and so on. Topology throws the distances away and keeps only one thing: a list of which sets count as open. It turns out that list is enough to talk about closeness, limits, continuity, connectedness, and compactness. This page sets up the list and the shortcuts for writing it down. Practice sits right after each idea, and the Mixed Practice at the bottom takes that hint away.
A topology is a collection of subsets that is closed under the operations we expect open sets to survive.
Let $X$ be a set. A topology on $X$ is a collection $\mathcal{T}$ of subsets of $X$ with the following properties:
Notice the asymmetry: any union, but only finite intersections. That's on purpose, and we'll see why in a second. A set together with a topology on it gets its own name.
A topological space $(X, \mathcal{T})$ is a pair consisting of a set $X$ and a topology $\mathcal{T}$ on $X$.
Even a two-point set already carries several different spaces.
List every topology on $X = \{a, b\}$.
Every topology contains $\emptyset$ and $X$, so the only choice is which of $\{a\}$ and $\{b\}$ to add. Any choice works, because unions and intersections of these sets land back in the list. So there are exactly four: $$\{\emptyset, X\}, \quad \{\emptyset, X, \{a\}\}, \quad \{\emptyset, X, \{b\}\}, \quad \{\emptyset, X, \{a\}, \{b\}\}.$$ The first is the trivial topology and the last is the discrete topology (both get names later on this page).
The members of $\mathcal{T}$ are the sets we call open.
Let $(X, \mathcal{T})$ be a topological space. A subset $U \subseteq X$ is an open set of $X$ if $U \in \mathcal{T}$.
That's the whole definition. "Open" doesn't mean anything until you say which topology you're using. Small examples make this concrete.
Let $X = \{a, b, c\}$. Which of these are topologies on $X$?
$$\mathcal{T}_1 = \{\emptyset, X, \{a\}, \{a, b\}\}, \qquad \mathcal{T}_2 = \{\emptyset, X, \{a\}, \{b\}\}.$$
$\mathcal{T}_1$ is a topology. It contains $\emptyset$ and $X$, and every union and intersection of its members is already in it: for example $\{a\} \cup \{a, b\} = \{a, b\}$ and $\{a\} \cap \{a, b\} = \{a\}$.
$\mathcal{T}_2$ is not. $\{a\}$ and $\{b\}$ are both in it, but $\{a\} \cup \{b\} = \{a, b\}$ is not.
On a finite set, checking "any union" only means checking unions of pairs, since every union of finitely many sets is built up two at a time. Build some topologies on $\{a, b, c\}$ yourself. There are exactly $29$ of them.
Your turn. One of these four collections is a topology; the other three each break an axiom the way people usually do.
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Now for why intersections have to be finite. Think about the usual open intervals of the real line, and watch what happens when you intersect more and more of them.
Every interval in that picture is open, and $0$ sits in all of them, but nothing else survives forever.
Each interval $\left(-\frac{1}{n}, \frac{1}{n}\right)$ is open in $\mathbb{R}$, but their intersection over all $n \in \mathbb{N}$ is $\{0\}$, which is not open. If the definition allowed infinite intersections, the usual open sets of $\mathbb{R}$ would not form a topology. So when you're checking the third axiom, check finite intersections only, and don't try to prove more than that.
One set can carry many topologies, and sometimes one topology contains another.
Suppose $\mathcal{T}$ and $\mathcal{T}'$ are two topologies on a given set $X$. If $\mathcal{T} \subseteq \mathcal{T}'$, we say that $\mathcal{T}'$ is finer than $\mathcal{T}$ (strictly finer if $\mathcal{T} \subsetneq \mathcal{T}'$).
Going the other way gives the opposite word.
Some books call the finer topology "stronger" and the coarser one "weaker." Other books use those two words the other way around. Nobody agreed. Stick with finer and coarser.
If $\mathcal{T} \subseteq \mathcal{T}'$, we also say $\mathcal{T}$ is coarser than $\mathcal{T}'$ (strictly coarser if $\mathcal{T} \subsetneq \mathcal{T}'$).
A picture that helps: a finer topology chops $X$ into more, smaller open pieces, like fine sand compared to coarse gravel. Sometimes neither topology contains the other.
Two topologies on $X$ are comparable if one of them is a subset of the other.
Here's an example of each situation.
On $X = \{a, b, c\}$, compare $\{\emptyset, X\}$, $\mathcal{T}_1 = \{\emptyset, X, \{a\}, \{a, b\}\}$, the collection of all subsets of $X$, and $\mathcal{T}_3 = \{\emptyset, X, \{b\}\}$.
$\{\emptyset, X\} \subseteq \mathcal{T}_1 \subseteq \{\text{all subsets of } X\}$, so these three are comparable, each strictly finer than the one before. But $\mathcal{T}_1$ and $\mathcal{T}_3$ are not comparable: $\{a\} \in \mathcal{T}_1 \setminus \mathcal{T}_3$ and $\{b\} \in \mathcal{T}_3 \setminus \mathcal{T}_1$.
Try one. Check the containment both ways before you answer.
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One way to build a coarser topology out of several: keep only the sets they all agree are open. Prove that this always works.
Let $X$ be a set and let $\{\mathcal{T}_\alpha\}_{\alpha \in A}$ denote the set of all topologies on $X$. Show that for any nonempty subcollection $B \subseteq A$, the set $\bigcap_{\alpha \in B} \mathcal{T}_\alpha$ is a topology on $X$.
Proof. Write $\mathcal{T} = \bigcap_{\alpha \in B} \mathcal{T}_\alpha$. We check the three conditions.
Unions are a different story.
The union of two topologies is usually not a topology. On $\{a, b, c\}$, $\{\emptyset, X, \{a\}\} \cup \{\emptyset, X, \{b\}\}$ contains $\{a\}$ and $\{b\}$ but not $\{a, b\}$. Don't assume an exercise about intersections works for unions too.
Here are all 29 topologies on $\{a, b, c\}$ at once, stacked from coarsest to finest. Pick one and see which others it can even be compared with.
Many times in topology, describing the entire topology $\mathcal{T}$ explicitly is too difficult. Try listing every open subset of $\mathbb{R}$! Instead we describe a smaller collection and let it generate the rest.
Let $X$ be a set. A basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called basis elements) such that:
The second condition is the one people forget, so here it is failing.
Is $\mathcal{B} = \{\{a\}, \{a, b\}, \{b, c\}\}$ a basis for a topology on $X = \{a, b, c\}$?
Condition 1 holds: every point lies in some member. But $b \in \{a, b\} \cap \{b, c\} = \{b\}$, and no member of $\mathcal{B}$ contains $b$ and fits inside $\{b\}$. So condition 2 fails, and $\mathcal{B}$ is not a basis. Adding $\{b\}$ to the collection fixes it.
Patch a broken collection yourself. Click sets in or out until both conditions hold; when one fails, the widget points at the exact point where it fails.
Now spot a basis without the widget's help. The wrong collections are the ones that look fine at a glance.
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A basis is a supply of "small open sets." The open sets of the topology are the ones you can fill in with them, point by point.
If $\mathcal{B}$ is a basis, the topology $\mathcal{T}$ generated by $\mathcal{B}$ is defined as follows: a subset $U \subseteq X$ is open in $X$ if for each $x \in U$, there exists a basis element $B \in \mathcal{B}$ such that $x \in B$ and $B \subseteq U$.
So "open" means "around every one of your points, there's room for a basis element." Calling this collection a topology is a promise, and the two basis conditions are exactly what it takes to keep it.
If $\mathcal{B}$ is a basis for a topology on $X$, then the collection $\mathcal{T}$ generated by $\mathcal{B}$ is a topology on $X$.
Proof. We check the three conditions.
Notice where each basis condition got used: the first one put $X$ in, and the second one kept intersections in. On a finite set you can check openness by hand.
Let $\mathcal{B} = \{\{a\}, \{c\}, \{a, b\}\}$ on $X = \{a, b, c\}$. Is $\{b, c\}$ open in the topology generated by $\mathcal{B}$? What about $\{a, c\}$?
$\{b, c\}$ is not open: the only basis element containing $b$ is $\{a, b\}$, and it doesn't fit inside $\{b, c\}$. But $\{a, c\}$ is open: $a \in \{a\} \subseteq \{a, c\}$ and $c \in \{c\} \subseteq \{a, c\}$. Checking every set this way gives the whole topology: $$\{\emptyset, \{a\}, \{c\}, \{a, b\}, \{a, c\}, X\}.$$
Build your own basis and watch its topology appear. Every set you add lights up the open sets it creates.
Your turn: test sets point by point.
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Every open set in that list is a union of basis elements ($\{a, c\} = \{a\} \cup \{c\}$). That's no accident, and the next theorem says so in general.
Let $X$ be a set and let $\mathcal{B}$ be a basis for the topology $\mathcal{T}$ on $X$. Then $\mathcal{T}$ equals the collection of all unions of elements of $\mathcal{B}$.
Proof. Each element of $\mathcal{B}$ is in $\mathcal{T}$ (for each of its points, it is a basis element around that point that fits inside itself), and since $\mathcal{T}$ is a topology, any union of them is in $\mathcal{T}$. Conversely, given $U \in \mathcal{T}$, choose for each $x \in U$ an element $B_x \in \mathcal{B}$ with $x \in B_x \subseteq U$ (it exists because $\mathcal{T}$ is generated by $\mathcal{B}$). Then $U = \bigcup_{x \in U} B_x$, so $U$ is a union of elements of $\mathcal{B}$. $\blacksquare$
That theorem turns "list the topology" into "list the unions." Count them for a fresh basis.
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Going the other direction: if you already have a topology, how can you tell whether some collection of open sets is a basis for it?
Let $(X, \mathcal{T})$ be a topological space. Suppose that $\mathcal{B}$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x \in U$, there exists $B \in \mathcal{B}$ such that $x \in B \subseteq U$. Then $\mathcal{B}$ is a basis for $\mathcal{T}$.
Proof. First, $\mathcal{B}$ is a basis. Given $x \in X$, the whole set $X$ is open, so by hypothesis there is $B \in \mathcal{B}$ with $x \in B \subseteq X$. Now let $x \in B_1 \cap B_2$ with $B_1, B_2 \in \mathcal{B}$. Since $B_1$ and $B_2$ are open, so is $B_1 \cap B_2$, and the hypothesis gives $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2$.
Next, $\mathcal{B}$ generates $\mathcal{T}$. Let $\mathcal{T}'$ be the topology generated by $\mathcal{B}$. If $U \in \mathcal{T}$ and $x \in U$, the hypothesis gives $B \in \mathcal{B}$ with $x \in B \subseteq U$, so $U \in \mathcal{T}'$ by definition. Conversely, if $V \in \mathcal{T}'$, then by the previous theorem $V$ is a union of elements of $\mathcal{B}$. Each element of $\mathcal{B}$ belongs to $\mathcal{T}$, and $\mathcal{T}$ is a topology, so $V \in \mathcal{T}$. $\blacksquare$
The most important basis of all is the one you've been using since precalculus.
If $\mathcal{B}$ is the collection of all open intervals in the real line, $(a, b) = \{x : a < x < b\}$, the topology generated by $\mathcal{B}$ is called the standard topology on the real line.
From now on, $\mathbb{R}$ means the real line with this topology unless we say otherwise. So in $\mathbb{R}$, a set is open exactly when every point of it has a little open interval around it that stays inside the set. Here's that sentence, and the theorem on unions, as a picture.
Near the edge of $U$ the intervals have to shrink, but they never run out, because the edge itself isn't in $U$. You don't even need every interval: the theorem on recognizing a basis shows that intervals with rational endpoints are enough.
Show that the countable collection $$\mathcal{B} = \{(a, b) : a < b, \; a, b \in \mathbb{Q}\}$$ is a basis that generates the standard topology on $\mathbb{R}$.
Proof. Every element of $\mathcal{B}$ is an open interval, so it is open in the standard topology. Let $U \subseteq \mathbb{R}$ be open in the standard topology and $x \in U$. There is an open interval around $x$ inside $U$, so there exists $\varepsilon > 0$ with $(x - \varepsilon, x + \varepsilon) \subseteq U$. Since the rationals are dense in $\mathbb{R}$, there exist $r_1, r_2 \in \mathbb{Q}$ with $r_1 \in (x - \varepsilon, x)$ and $r_2 \in (x, x + \varepsilon)$. Then $(r_1, r_2) \in \mathcal{B}$ and $$x \in (r_1, r_2) \subseteq (x - \varepsilon, x + \varepsilon) \subseteq U.$$ By the theorem on recognizing a basis, $\mathcal{B}$ is a basis for the standard topology on $\mathbb{R}$. $\blacksquare$
With bases in hand, comparing two topologies becomes a test you can run on basis elements alone.
Let $\mathcal{B}$ and $\mathcal{B}'$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}'$, respectively, on $X$. Then the following are equivalent:
Proof. $(2) \implies (1)$. Given $U \in \mathcal{T}$, we show $U \in \mathcal{T}'$. Let $x \in U$. Since $\mathcal{B}$ generates $\mathcal{T}$, there is $B \in \mathcal{B}$ with $x \in B \subseteq U$. By condition (2), there is $B' \in \mathcal{B}'$ with $x \in B' \subseteq B$. Then $x \in B' \subseteq U$, so $U \in \mathcal{T}'$ by definition.
$(1) \implies (2)$. We are given $x \in X$ and $B \in \mathcal{B}$ with $x \in B$. By definition $B \in \mathcal{T}$, and $\mathcal{T} \subseteq \mathcal{T}'$ by condition (1), so $B \in \mathcal{T}'$. Since $\mathcal{T}'$ is generated by $\mathcal{B}'$, there exists $B' \in \mathcal{B}'$ with $x \in B' \subseteq B$. $\blacksquare$
Read condition (2) as a game: someone hands you a point $x$ and a "big" basis element $B$ around it, and you have to fit one of your basis elements around $x$ inside $B$. If you can always win, your topology is finer. Here's the game played between open intervals $(a, b)$ and half-open intervals $[a, b)$ (we'll name that second topology in a moment).
In case 1 you can always win with $[x, b)$. In case 2 you can never win, because the point $x$ sits right at the closed end. The order matters: "finer" is a one-way street.
Don't mix up the two roles in condition (2). The basis element you're given comes from the coarser topology, and the one you find comes from the finer topology. A good sanity check: the finer topology has more open sets, so it's the one with smaller pieces to fit inside things.
Play the game once more, on paper this time. Sometimes the honest answer is that nothing fits.
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Sometimes even a basis is more than you need, and a smaller collection still generates everything.
A subbasis $\mathcal{S}$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$.
The only requirement is that the pieces cover $X$. They don't have to fit together the way basis elements do, because the next step repairs that.
The topology generated by the subbasis $\mathcal{S}$ is the collection $\mathcal{T}$ of all unions of finite intersections of elements of $\mathcal{S}$.
The finite intersections of elements of $\mathcal{S}$ form a basis, and then the unions fill in the topology. (Proving that this really is a topology is the Spicy Bonus at the end.)
Subbases sound like a technicality, but they're exactly how the product topology on an infinite product gets defined: the subbasis elements restrict one coordinate at a time.
On a finite set you can watch both passes happen.
Find the topology on $X = \{a, b, c\}$ generated by the subbasis $\mathcal{S} = \{\{a, b\}, \{b, c\}\}$.
First the finite intersections: $\{a, b\}$, $\{b, c\}$, and $\{a, b\} \cap \{b, c\} = \{b\}$. These form a basis. Then every union: $\{a, b\} \cup \{b, c\} = X$, and the rest are already listed. With $\emptyset$, the topology is $$\{\emptyset, X, \{b\}, \{a, b\}, \{b, c\}\}.$$
Here are both passes on four points, one new set at a time. Each loop starts from a different subbasis.
Notice the middle column: the intersections are what turn a subbasis into a basis. Now run both passes yourself and count.
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The same two passes work on infinite sets too.
Let $\mathcal{S}$ be the collection of all open rays $(-\infty, b)$ and $(a, \infty)$ in $\mathbb{R}$. What topology does $\mathcal{S}$ generate?
The union of the rays is $\mathbb{R}$, so $\mathcal{S}$ is a subbasis. Finite intersections of rays include $(a, \infty) \cap (-\infty, b) = (a, b)$, so every open interval is in the generated topology, and so is every union of open intervals. Conversely, every finite intersection of rays is a ray, an interval, $\mathbb{R}$, or $\emptyset$, and all of those are open in the standard topology. So $\mathcal{S}$ generates exactly the standard topology.
The whole example in one picture: two rays, each missing half the line, overlap in an interval.
This section is simply a list of (some of) the topologies that topologists run into from time to time. The more important ones get pages of their own later. First, the two extremes, which every set has.
Let $X$ be a set. The trivial topology is the topology which only contains $\emptyset$ and $X$.
That's the coarsest topology possible. At the opposite end, everything is open.
Let $X$ be a set. The collection of all subsets of $X$ is a topology on $X$, called the discrete topology.
Every topology on $X$ sits between these two. Next, a topology where open sets are enormous.
Let $X$ be a set. The finite complement topology, denoted $\mathcal{T}_f$, is the collection of all subsets $U \subseteq X$ such that $X \setminus U$ is finite or $U = \emptyset$.
Here "finite" really means finite, and it changes which sets count.
In $(\mathbb{R}, \mathcal{T}_f)$, which of these are open: $(0, 1)$, $\mathbb{R} \setminus \{0, 5\}$, $\mathbb{R} \setminus \mathbb{Z}$?
Only $\mathbb{R} \setminus \{0, 5\}$, whose complement $\{0, 5\}$ is finite. The complement of $(0, 1)$ is infinite, so this interval, open in the standard topology, is not open here. The complement of $\mathbb{R} \setminus \mathbb{Z}$ is $\mathbb{Z}$, also infinite.
In $(\mathbb{R}, \mathcal{T}_f)$, an open set is "all of $\mathbb{R}$ except finitely many points." Any two nonempty open sets overlap (each is missing only finitely many points), which makes this space a gold mine of counterexamples.
However many points the two sets throw away, the line has infinitely many left, so they always share some. Now we can line up the topologies so far.
Show that on $\mathbb{R}$, $$\{\emptyset, \mathbb{R}\} \subsetneq \mathcal{T}_f \subsetneq \{\text{standard}\} \subsetneq \{\text{all subsets of } \mathbb{R}\},$$ so each of the trivial, finite complement, standard, and discrete topologies is strictly finer than the one before.
The trivial topology is inside every topology, and $\mathbb{R} \setminus \{0\}$ is in $\mathcal{T}_f$ but not in the trivial topology.
Next, let $U \in \mathcal{T}_f$ be nonempty, so $U = \mathbb{R} \setminus \{p_1, \dots, p_k\}$ with $p_1 < \dots < p_k$. Then $U$ is the union of $(-\infty, p_1)$, the intervals $(p_i, p_{i+1})$, and $(p_k, \infty)$, and each ray is a union of open intervals, such as $(-\infty, p_1) = \bigcup_{n \in \mathbb{N}} (p_1 - n, p_1)$. So $U$ is open in the standard topology. But $(0, 1)$ is standard-open and its complement is infinite, so it is not in $\mathcal{T}_f$.
Last, every set is open in the discrete topology, but $\{0\}$ contains no open interval, so it isn't standard-open. Every inclusion is strict.
Because nonempty open sets in $(\mathbb{R}, \mathcal{T}_f)$ always overlap, you can't put two points in disjoint open sets. On the Hausdorff page we'll see that this lets a sequence of distinct points converge to every real number at once.
What if "finite" becomes "countable"? It's still a topology. Prove it, and remember to check finite intersections only.
Let $X$ be a set and let $\mathcal{T}_c$ be the collection of all subsets $U \subseteq X$ such that $X \setminus U$ is either countable or all of $X$. Show that $\mathcal{T}_c$ is a topology on $X$.
Proof. We check the three conditions.
The next one looks almost the same, but changes "countable" to "infinite."
Let $X$ be a set. Is the collection $$\mathcal{T}_\infty = \{U \subseteq X : X \setminus U \text{ is infinite, empty, or all of } X\}$$ a topology on $X$?
It depends on $X$. If $X$ is finite, no complement is infinite, so $\mathcal{T}_\infty = \{\emptyset, X\}$, the trivial topology, which is a topology.
If $X$ is infinite, it never is. Pick a point $p \in X$. For each $x \neq p$, the complement $X \setminus \{x\}$ is infinite, so $\{x\} \in \mathcal{T}_\infty$. But their union $$\bigcup_{x \neq p} \{x\} = X \setminus \{p\}$$ has complement $\{p\}$, which is not infinite, not empty, and not all of $X$ (since $X$ has more than one point). So $\mathcal{T}_\infty$ is not closed under unions. On $\mathbb{N}$ with $p = 1$, that union is $\bigcup_{k=2}^{\infty} \{k\} = \mathbb{N} \setminus \{1\}$.
Now three topologies on $\mathbb{R}$ that are built from bases. The standard topology (above) uses open intervals. Change the intervals and you change the topology.
Let $\mathcal{B}$ be the collection of all half-open intervals of the form $[a, b) = \{x : a \le x < b\}$ with $a < b$. The topology generated by $\mathcal{B}$ is called the lower limit topology on $\mathbb{R}$. We write $\mathbb{R}_l$ for $\mathbb{R}$ with this topology.
The widget above showed that $\mathbb{R}_l$ is strictly finer than the standard topology. Which end of the interval is closed matters.
Are $[0, 1)$ and $(0, 1]$ open in $\mathbb{R}_l$?
$[0, 1)$ is open: it is a basis element. $(0, 1]$ is not: every basis element around $1$ has the form $[c, d)$ with $c \le 1 < d$, so it contains points bigger than $1$, which are not in $(0, 1]$.
The next topology punches holes in the standard intervals instead.
Let $K$ denote the set of all numbers of the form $\frac{1}{n}$ for $n \in \mathbb{N}$, and let $\mathcal{B}$ be the collection of all open intervals $(a, b)$, along with the sets of the form $(a, b) \setminus K$. The topology generated by $\mathcal{B}$ is called the K-topology on $\mathbb{R}$. When $\mathbb{R}$ is given this topology, it is denoted $\mathbb{R}_K$.
$\mathbb{R}_K$ is also strictly finer than the standard topology: $(-1, 1) \setminus K$ is open in $\mathbb{R}_K$, but no open interval around $0$ fits inside it, because every such interval contains some $\frac{1}{n}$. Zoom in on $0$ and watch it happen.
The K-topology exists almost entirely to be a counterexample. It's the topology equivalent of a crash-test dummy.
That makes six topologies on $\mathbb{R}$ so far. Pick any two and the ladder tells you how they compare, with a set that proves it.
Test the zoo yourself. Pick a topology and a set, then drag the probe through the set: it's open exactly when every point has a basis element around it that fits.
Now sort a whole handful at once. Drop each set in the bin that says which of the four topologies it's open in.
One more: which topology makes this set open?
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The standard topology on $\mathbb{R}$ only used the order $<$, never distances. That works on any ordered set.
Let $X$ be a set with a simple order relation, and assume $X$ has more than one element. Let $\mathcal{B}$ be the collection of all sets of the following types:
The order topology on $X$ is the topology generated by $\mathcal{B}$.
That $\mathcal{B}$ really is a basis: every point lies in one of these intervals (that's where we need two elements), and two of them overlap in another set of the same kinds, or not at all. The half-open intervals at the ends are there so the smallest and largest elements (if they exist) still have basis elements around them.
What is the order topology on $\mathbb{N} = \{1, 2, 3, \dots\}$ with its usual order?
Every one-point set is a basis element: $\{n\} = (n - 1, n + 1)$ for $n \ge 2$, and $\{1\} = [1, 2)$ (here $1$ is the smallest element). Every subset is a union of one-point sets, so every subset is open. The order topology on $\mathbb{N}$ is the discrete topology.
And for $\mathbb{R}$ with its usual order, the order topology is the standard topology: $\mathbb{R}$ has no smallest or largest element, so the basis is exactly the open intervals. There's one more topology on the list, and it's important enough that it gets a full page.
The product topology on $X \times Y$ uses the "rectangles" $U \times V$ as its basis, with $U$ open in $X$ and $V$ open in $Y$. In $\mathbb{R} \times \mathbb{R}$ those are the open rectangles of the plane.
Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim 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 vocabulary. Each blank is a word from a definition somewhere on this page.
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Then the theorems and facts, with one piece missing.
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Next, a proof from this page with one line to fill in. Every wrong line is false or doesn't follow.
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Next, 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; a few types need their answer choices, and they'll say so.
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Last, the Spicy Bonus. The definition of a subbasis promised that "all unions of finite intersections" is a topology. Time to prove it.
Let $\mathcal{S}$ be a subbasis for a topology on $X$. Prove that the collection $\mathcal{B}$ of all finite intersections of elements of $\mathcal{S}$ is a basis, and conclude that the collection of all unions of elements of $\mathcal{B}$ is a topology on $X$.
Proof. $\mathcal{B}$ is a basis. Given $x \in X$, the union of the elements of $\mathcal{S}$ is $X$, so $x$ lies in some $S \in \mathcal{S}$, and $S$ (an intersection of one element) is in $\mathcal{B}$. For the second condition, let $B_1 = S_1 \cap \dots \cap S_m$ and $B_2 = S_1' \cap \dots \cap S_k'$ be elements of $\mathcal{B}$. Then $$B_1 \cap B_2 = S_1 \cap \dots \cap S_m \cap S_1' \cap \dots \cap S_k'$$ is again a finite intersection of elements of $\mathcal{S}$, so it is in $\mathcal{B}$. For $x \in B_1 \cap B_2$ we can take $B_3 = B_1 \cap B_2$.
The unions form a topology. Let $\mathcal{T}$ be the topology generated by the basis $\mathcal{B}$; it is a topology by the theorem that a basis generates a topology. By the theorem on unions of basis elements, $\mathcal{T}$ is exactly the collection of all unions of elements of $\mathcal{B}$. That is the collection of all unions of finite intersections of elements of $\mathcal{S}$, so this collection is a topology. $\blacksquare$