In metric spaces, sequences can do everything: detect closures, test continuity, characterize compactness. In general topological spaces they can't, because a point might need "more than countably many" shrinking neighborhoods to be reached. Nets fix this by replacing the index set $\mathbb{N}$ with something more flexible. With nets, every one of those sequence theorems becomes true in every space.
What made $\mathbb{N}$ a good index set for sequences was its order: "later" terms. We keep an order, but allow it to branch.
A relation on a set $A$, denoted $\preceq$, is called a partial order if the following conditions hold:
In a partial order two elements can be incomparable (neither is $\preceq$ the other). What we need is that any two elements are eventually "overtaken" together.
A directed set $\mathcal{D}$ is a set with a partial order $\preceq$ such that for any $\alpha, \beta \in \mathcal{D}$, there exists $\gamma \in \mathcal{D}$ with $\alpha \preceq \gamma$ and $\beta \preceq \gamma$.
Click pairs of elements in a few partial orders and look for a common upper bound.
Here's the directed set that matters most in topology.
Let $x$ be a point of a space $X$, and let $\mathcal{D}$ be the set of all neighborhoods of $x$, with $U \preceq V$ meaning $V \subset U$. Is $\mathcal{D}$ directed?
Yes. "Bigger in the order" means "smaller as a set," so moving up the order means closing in on $x$. Given neighborhoods $U$ and $V$, the set $W = U \cap V$ is a neighborhood of $x$, and $W \subset U$, $W \subset V$, so $U \preceq W$ and $V \preceq W$.
In a metric space we could just use the balls $B(x, \frac{1}{n})$, which are indexed by $\mathbb{N}$. That's why the sequence lemma worked there. In general there may be no countable list that does the job, so we use all the neighborhoods.
A sequence is a function from $\mathbb{N}$. A net is a function from any directed set.
A net in a topological space $X$ is a function $\Gamma: \mathcal{D} \to X$ from a directed set $\mathcal{D}$ into $X$.
A net is what a sequence becomes after grad school: same job, but now it insists on being indexed by an arbitrary directed set, and it has opinions about it.
The two ways a net can "be in" a set generalize "for all large $n$" and "for infinitely many $n$."
Let $\Gamma: \mathcal{D} \to X$ be a net and $A \subset X$. $\Gamma$ is eventually in $A$ if there exists $\alpha \in \mathcal{D}$ such that $\Gamma(\beta) \in A$ for all $\beta$ with $\alpha \preceq \beta$.
The weaker version only asks that you can always find a later term in $A$.
$\Gamma$ is frequently in $A$ if for each $\alpha \in \mathcal{D}$ there exists $\beta$ with $\alpha \preceq \beta$ and $\Gamma(\beta) \in A$.
Every sequence is a net on $\mathbb{N}$, so we can see these two definitions on sequences first.
Convergence is "eventually in every neighborhood," exactly as for sequences.
A net $\Gamma: \mathcal{D} \to X$ in a topological space $X$ is said to converge to $x \in X$ if for every neighborhood $U$ of $x$, $\Gamma$ is eventually in $U$.
With the neighborhood directed set from the example above, if every neighborhood $U$ of $x$ meets a set $A$, we can choose a point $\Gamma(U) \in U \cap A$. That net converges to $x$: once $U \subset W$, the point $\Gamma(U)$ is in $W$. This is the net version of the sequence lemma, and it's Exercise 1.
For sequences, Hausdorff spaces gave unique limits but not the other way around. For nets the two are equivalent.
A topological space $X$ is Hausdorff if and only if every convergent net in $X$ has exactly one limit.
Proof. Suppose $X$ is Hausdorff and a net $\Gamma: \mathcal{D} \to X$ converges to $x$ and to $y$ with $x \neq y$. Take disjoint neighborhoods $U \ni x$ and $V \ni y$. By convergence there are $\alpha_x, \alpha_y \in \mathcal{D}$ with $\Gamma(\beta) \in U$ whenever $\alpha_x \preceq \beta$, and $\Gamma(\beta) \in V$ whenever $\alpha_y \preceq \beta$. Since $\mathcal{D}$ is directed, some $\gamma$ has $\alpha_x \preceq \gamma$ and $\alpha_y \preceq \gamma$, so $\Gamma(\gamma) \in U \cap V = \emptyset$, which is impossible. So $x = y$.
Conversely, suppose $X$ is not Hausdorff: there are points $p \neq q$ such that every neighborhood $U$ of $p$ meets every neighborhood $V$ of $q$. Let $\mathcal{D}$ be the set of pairs $(U, V)$ with $U$ a neighborhood of $p$ and $V$ a neighborhood of $q$, ordered by $(A, B) \preceq (U, V)$ when $U \subset A$ and $V \subset B$. It is directed: $(U \cap U', V \cap V')$ is above both $(U, V)$ and $(U', V')$. For each $(U, V)$ choose $\Gamma(U, V) \in U \cap V$. Given a neighborhood $W$ of $p$, set $\alpha = (W, X)$; if $\alpha \preceq (A, B)$ then $A \subset W$, so $\Gamma(A, B) \in A \cap B \subset W$. So $\Gamma$ is eventually in $W$, and $\Gamma \to p$. In the same way $\Gamma \to q$. So $\Gamma$ is a convergent net with two different limits. $\blacksquare$
And continuity is exactly "preserves limits of nets."
A function $f: X \to Y$ between topological spaces is continuous if and only if for every net $\Gamma$ converging to $x \in X$, the net $f \circ \Gamma$ in $Y$ converges to $f(x)$.
Proof. Suppose $f$ is continuous and $\Gamma \to x$. Let $V$ be a neighborhood of $f(x)$. Then $f^{-1}(V)$ is a neighborhood of $x$, so $\Gamma$ is eventually in $f^{-1}(V)$, which means $f \circ \Gamma$ is eventually in $V$. So $f \circ \Gamma \to f(x)$.
Conversely, suppose $f$ is not continuous. Then there is an open $V \subset Y$ such that $A = f^{-1}(V)$ is not open, so some $x \in A$ is not in $\operatorname{int}(A)$: every neighborhood $W$ of $x$ contains a point outside $A$. Let $\mathcal{D}$ be the neighborhoods of $x$, ordered by reverse inclusion, and for each $W \in \mathcal{D}$ choose $\Gamma(W) \in W \setminus A$. Then $\Gamma \to x$ (once $W$ is inside a given neighborhood $U$, so is $\Gamma(W)$). But $f(\Gamma(W)) \notin V$ for every $W$, while $V$ is a neighborhood of $f(x)$ (since $x \in A$). So $f \circ \Gamma$ is never in $V$, and it does not converge to $f(x)$. $\blacksquare$
The Hausdorff page warned that sequences can't detect the Hausdorff property both ways. The pairs-of-neighborhoods net above is exactly the kind of net no sequence can imitate.
For subsequences we picked increasing indices $n_1 < n_2 < \cdots$. For subnets we only need a map between directed sets that keeps going "out to the end."
Let $\mathcal{D}$ and $\tilde{\mathcal{D}}$ be directed sets and let $h: \tilde{\mathcal{D}} \to \mathcal{D}$ be a function. We say that $h$ is final if for every $\alpha \in \mathcal{D}$ there exists $\tilde{\alpha} \in \tilde{\mathcal{D}}$ such that $\alpha \preceq h(\tilde{\beta})$ whenever $\tilde{\alpha} \preceq \tilde{\beta}$.
Composing with a final map gives a subnet.
A subnet of a net $\Gamma: \mathcal{D} \to X$ is a composition $\Gamma \circ h$, where $h: \tilde{\mathcal{D}} \to \mathcal{D}$ is a final function.
A subsequence $(x_{n_k})$ is a subnet, with $h(k) = n_k$: since $n_k \ge k$, the terms go past any index $N$ once $k \ge N$. But subnets can also repeat terms or be indexed by much bigger sets, and that flexibility is what makes the next idea possible.
A net in $X$ is called universal if for every $A \subset X$, the net is either eventually in $A$ or eventually in $X \setminus A$.
A universal net has made up its mind about every subset of $X$. The sequence $(-1)^n + \frac{1}{n}$ from the widget is not universal: it is frequently in $A = [0.75, 1.25]$ and frequently outside it.
Every net has a universal subnet.
(The proof needs Zorn's lemma; it's on the Tychonoff's Theorem page.) Universal nets give one of four equivalent ways to say "compact."
Let $X$ be a topological space. The following are equivalent:
(The equivalence of (1), (3), and (4) is proved on Tychonoff's Theorem; (1) and (2) are the complement trick from Compact Spaces.) Compare (4) with sequential compactness on the Compact Spaces page: replace "sequence" by "net" and "subsequence" by "subnet," and the statement becomes true in every space.
When a sequence argument from analysis seems to fail in a general space, don't throw the argument away. Rewrite it with nets: index by the neighborhoods of the point instead of by $n$, and replace "for all $n \ge N$" with "for all $\beta \succeq \alpha$." The rest of the proof usually goes through word for word.
The net version of the sequence lemma, with no metric needed.
Let $X$ be a topological space and $A \subset X$. Prove that $p \in \overline{A}$ if and only if there exists a net $\Gamma$ in $A$ that converges to $p$.
Proof. Suppose $p \in \overline{A}$. Let $\mathcal{U}_p$ be the set of neighborhoods of $p$, ordered by $U \preceq V$ if and only if $V \subset U$; this is a directed set. For each $W \in \mathcal{U}_p$, the set $W \cap A$ is nonempty (since $p \in \overline{A}$), so choose $a_W \in W \cap A$. The net $(a_W)_{W \in \mathcal{U}_p}$ lies in $A$ and converges to $p$: given $B \in \mathcal{U}_p$, if $B \preceq C$ then $a_C \in C \cap A \subset B$, so the net is eventually in $B$.
Conversely, suppose a net $(x_\alpha)_{\alpha \in \mathcal{D}}$ in $A$ converges to $p$. Then for every neighborhood $U$ of $p$, the net is eventually in $U$, so in particular $U$ contains some $x_\alpha \in A$. Every neighborhood of $p$ meets $A$, so $p \in \overline{A}$. $\blacksquare$
A prelim problem where nets make a short proof possible. Follow the hint.
Let $X$ be Hausdorff and $f: X \to X$ continuous with $f \circ f = f$. Prove that $f(X)$ is closed. (Hint: assume $f(X)$ is not closed, and construct a net that violates the unique limits theorem.)
Proof. Suppose $f(X)$ is not closed. Then there is a point $x \in \overline{f(X)} \setminus f(X)$. By Exercise 1, there is a net $(\Gamma_\alpha)_{\alpha \in \mathcal{D}}$ in $f(X)$ converging to $x$. Each $\Gamma_\alpha = f(x_\alpha)$ for some $x_\alpha \in X$, so $$f(\Gamma_\alpha) = f(f(x_\alpha)) = f(x_\alpha) = \Gamma_\alpha.$$ So $f \circ \Gamma = \Gamma$. Since $f$ is continuous, $f \circ \Gamma \to f(x)$; that is, $\Gamma \to f(x)$. But also $\Gamma \to x$. Since $X$ is Hausdorff, limits of nets are unique, so $f(x) = x$, which puts $x \in f(X)$. That contradicts the choice of $x$. Hence $f(X)$ is closed. $\blacksquare$