# General Topology/Nets

Definition (net):
Let $X$ be a topological space, and let $(\Lambda ,\leq )$ be a directed set. A net is a family $(x_{\lambda })_{\lambda \in \Lambda }$.
Definition (net convergence):
Let $X$ be a topological space, let $(x_{\lambda })_{\lambda \in \Lambda }$ be a net (where then of course $\Lambda ,\leq )$ is a directed set) and let $x\in X$. The sequence $(x_{\lambda })_{\lambda \in \Lambda }$ is said to converge to $x$ if and only if for every neighbourhood $M\in N(x)$ there exists $\eta =\eta (M)\in \Lambda$ so that for all $\lambda \geq \eta$ we have $x_{\lambda }\in M$. For this we write $\lim _{\lambda \in \Lambda }x_{\lambda }=x$.
As are filters, nets are analogues of sequences, which are used to adapt theorems which otherwise would only hold for "nice" spaces to the setting of general topological spaces. The downside is, as with filters, that theorems involving nets often use the axiom of choice. Whether you use nets or filters is a matter of taste, and a matter of selecting that tool which uses the least amount of choice in your given situation.
Proposition (closedness is equivalent to containing all net limits):
Let $X$ be a topological space. Then a set $A\subseteq X$ is closed if and only if for all nets $(x_{\lambda })_{\lambda \in \Lambda }$ so that $\forall \lambda \in \Lambda :x_{\lambda }\in A$ which converge to a point $x\in X$, we have $x\in A$.
Proof: Suppose first that $A$ is closed, and that $(x_{\lambda })_{\lambda \in \Lambda }$ is a convergent net in $A$ to a limit $x$. Suppose $x\notin A$. Then for all $M\subseteq N(x)$, we find $\eta \in \Lambda$ so that at least $x_{\eta }\in M$, which implies that $M\cap A\neq \emptyset$, and $x\in \partial A$ since $x\in M$, $x\notin A$. Since a set is closed if and only if it contains its boundary, we conclude that $x\in A$, a contradiction. Suppose then that $A$ contains the limit of all its convergent nets. Then let $x\in \partial A$, and note that $N(x)$ is a directed set, ordered by inclusion. For each $M\in N(x)$, choose $x_{M}\in A\cap M$ by definition of $\partial A$, and observe that $(x_{M})_{M\in N(X)}$ forms a net convergent to $x$. Thus, $x\in A$, and since $x$ was arbitrary, $\partial A\subseteq A$ and $A$ is closed. $\Box$
Proposition (characterisation of continuity by net convergence):
Let $X,Y$ be topological spaces and $f:X\to Y$ a function. Then $f$ is continuous if and only if for all nets $(x_{\lambda })_{\lambda \in \Lambda }$ in $X$ that converge to a certain point $x_{\infty }\in X$, the net $(f(x_{\lambda }))_{\lambda \in \Lambda }$ converges to $f(x_{\infty })$.
Proof: Suppose first that $f:X\to Y$ is continuous, and suppose that $(x_{\lambda })_{\lambda \in \Lambda }$ converges to $x_{\infty }\in X$. Then let $V$ be any open neighbourhood of $f(x_{\infty })$, and set $U:=f^{-1}(V)$, which is open by the continuity of $f$. Then pick $\mu \in \Lambda$ so that for all $\lambda \geq \mu$ we have $x_{\lambda }\in U$; for such $\lambda$ we will then have $f(x_{\lambda })\in V$ by definition of the preimage.
Suppose now that $f:X\to Y$ has the property that net convergence $(x_{\lambda })_{\lambda \in \Lambda }$ to $x_{\infty }$ implies net convergence of $(f(x_{\lambda }))_{\lambda \in \Lambda }$ to $f(x_{\infty })$. Then suppose that $A\subseteq Y$ is closed. If we prove that $f^{-1}(A)$ is closed, we've proven the theorem. It suffices to prove that $f^{-1}(A)$ contains all limits of its nets. So suppose that $(x_{\lambda })_{\lambda \in \Lambda }$ is a net in $f^{-1}(A)$ that converges to some $x_{\infty }\in X$. Then $f(x_{\infty })\in A$ and thus $x_{\infty }\in f^{-1}(A)$, so that $f^{-1}(A)$ is indeed closed. $\Box$
Definition (subnet):
Let $X$ be a topological space and $(x_{\lambda })_{\lambda \in \Lambda }$ be a net in it. A subnet of $(x_{\lambda })_{\lambda \in \Lambda }$ is a net $(y_{\sigma })_{\sigma \in \Sigma }$ together with a final order homomorphism $f:\Sigma \to \Lambda$ so that $\forall \sigma \in \Sigma :x_{f(\sigma )}=x_{\sigma }$.
Proposition (compactness is equivalent to the existence of convergent subnets):
Let $X$ be a topological space, and let $A\subseteq X$ be a subset. $A$ is compact if and only if for all nets $(x_{\lambda })_{\lambda \in \Lambda }$ in $A$, there exists $x\in A$ and a subnet $(y_{\sigma })_{\sigma \in \Sigma }$ of $(x_{\lambda })_{\lambda \in \Lambda }$ so that $\lim _{\sigma \in \Sigma }y_{\sigma }=x$.
Proof: Suppose first that $A$ is compact, and let $(x_{\lambda })_{\lambda \in \Lambda }$ be a net in $A$. Suppose that $(x_{\lambda })_{\lambda \in \Lambda }$ does not have a convergent subnet. Then for each $x\in A$, we find an open neighbourhood $U_{x}$ of $x$ and a $\lambda _{x}$ such that for all $\lambda \geq \lambda _{x}$, we have $x_{\lambda }\notin U_{x}$; for otherwise, if $x$ is a point so that for all open neighbourhoods $U_{x}$ and $\lambda \in \Lambda$ there exists $\mu \in \Lambda$ so that $x_{\mu }\in U_{x}$, then we note that the set of all $U_{x}$ (that is, the sets of all open neighbourhoods of $x$) forms a directed set when ordered inversely by inclusion (which shall henceforth be denoted by $\Gamma$). Note that we may take the product order on $\Lambda \times \Gamma$, and then we may define an order morphism $f:\Lambda \times \Gamma \to \Lambda$ by sending $(\lambda ,\gamma )\mapsto \lambda$. Consider the subset $S\subseteq \Lambda \times \Gamma$ so that $x_{\lambda }\in \gamma$. We define a subnet of $(x_{\lambda })_{\lambda \in \Lambda }$ as thus: Define $(y_{(\lambda ,\gamma )})_{(\lambda ,\gamma )\in S}$ by $y_{(\lambda ,\gamma )}:=x_{\lambda }$, and the order morphism induced by $f$, which is final since for any $\lambda _{0}$, we find $\lambda \geq \lambda _{0}$ so that $x_{\lambda }\in \gamma _{0}$, when $(\lambda _{0},\gamma _{0})\in T$ is arbitrary. Now $(y_{(\lambda ,\gamma )})_{(\lambda ,\gamma )\in S}$ converges to $x$, a contradiction. But then we apply compactness to the $U_{x}$, so that we gain points $x_{1},\ldots ,x_{n}$ so that $U_{x_{1}}\cup \cdots \cup U_{x_{n}}$ covers $A$. Then we choose an upper bound $\lambda _{0}$ of $\lambda _{x_{1}},\ldots ,\lambda _{x_{n}}$ by successively choosing an upper bound of $\lambda _{1},\lambda _{2}$, then an upper bound of that upper bound and $\lambda _{3}$ and so on. Then for $\lambda \geq \lambda _{0}$, $x_{\lambda }$ is not contained in any of the $U_{x_{j}}$, so that in particular $x_{\lambda _{0}}\notin A$, a contradiction. Conversely, if every net contains a convergent subnet, let $(U_{\alpha })_{\alpha \in A}$ be an open cover of $X$. Suppose that it doesn't admit any finite subcover, and consider the set of all finite subfamilies of $(U_{\alpha })_{\alpha \in A}$ ordered by inclusion, which is a directed set. Then define a net indexed over that set by choosing, for each finite index set $\alpha _{1},\ldots ,\alpha _{n}$, the element $x_{U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}}\in X\setminus (U_{\alpha _{1}}\cup \cdots \cup U_{\alpha _{n}})$. By assumption, upon defining $\Lambda$ to be the set of all finite subfamilies of $(U_{\alpha })_{\alpha \in A}$, we find that $(x_{\lambda })_{\lambda \in \Lambda }$ has a convergent subnet $(y_{\gamma })_{\gamma \in \Gamma }$. Let $y$ be the point to which this subnet converges, and since $(U_{\alpha })_{\alpha \in A}$ is a cover, pick $\alpha _{0}\in A$ so that $y\in U_{\alpha _{0}}$. Note that $(U_{\alpha _{0}})$ is a finite subfamily of $(U_{\alpha })_{\alpha \in A}$, so that by finality of the order homomorphism $f:\Gamma \to A$ we find $\gamma _{1}\in \Gamma$ so that $f(\gamma )$ contains $U_{\alpha _{0}}$. On the other hand, since $(y_{\gamma })_{\gamma \in \Gamma }$ converges to $y$, we find $\gamma _{2}\in \Gamma$ so that for $\gamma \geq \gamma _{2}$ we have $x_{\gamma }\in U_{\alpha _{0}}$. Finally, since $\Gamma$ is directed, let $\gamma _{3}$ be an upper bound for $\gamma _{1}$ and $\gamma _{2}$, then $f(\gamma _{3})$ contains $U_{\alpha _{0}}$, but then $y_{\gamma _{3}}=x_{f(\gamma _{3})}$ is not contained in $U_{\alpha _{0}}$, a contradiction. $\Box$
Definition (sequence):
Let $X$ be a topological space. A sequence is a net indexed over the directed set of the natural numbers.
In order to denote that a sequence $(x_{n})_{n\in \mathbb {N} }$ converges to a point $x$, we shall resort to the useful notation $x_{n}\to x$.
Definition (sequentially closed):
Let $X$ be a topological space. A subset $A\subseteq X$ is called sequentially closed iff for all sequences $(x_{n})_{n\in \mathbb {N} }$ in $A$ that converge in $X$, the limits are all contained within $A$.
Proposition (characterisation of closedness by sequences in first-countable spaces):
Let $X$ be a first-countable topological space. Then a subset $A\subseteq X$ is closed iff it is sequentially closed.
Proof: Suppose first that $A$ is closed. Then certainly every limit of a convergent sequence in $A$ is contained within $A$, because a sequence is a net, and the corresponding statement for nets is true without the axiom of choice. Suppose now that $A$ is sequentially closed. Then let $x\in \partial A$; we are to show that $x\in A$. Choose a countable basis $(U_{n})_{n\in \mathbb {N} }$ for the neighbourhood filter $N(x)$. By the axiom of countable choice, pick $x_{n}\in U_{1}\cap \cdots \cap U_{n}\cap A$ for each $n\in \mathbb {N}$. Then the sequence $(x_{n})_{n\in \mathbb {N} }$ converges to $x$, and hence $x\in A$. $\Box$
Definition (sequential continuity):
A function $f:X\to Y$ between topological spaces is called sequentially continuous iff for all sequences $(x_{n})_{n\in \mathbb {N} }$ in $X$ that converge to some $x\in X$, we have $f(x_{n})\to f(x)$.
Proposition (sequence criterion for continuity in first countable spaces):
Let $X$ be a first-countable space, $Y$ a topological space and $f:X\to Y$ a function. Then $f$ is continuous if and only if it is sequentially continuous.
Proof: If $f$ is continuous, certainly $f(x_{n})\to f(x)$ whenever $x_{n}\to x$, since a neighbourhood $V$ of $f(x)$ gives rise to a neighbourhood $f^{-1}(V)$ of $x$, into which $x_{n}$ will be for $n$ sufficiently large, and then $f(x_{n})\in V$ by def. of the preimage.
If $f$ is sequentially continuous, let $A\subseteq Y$ be closed, and set $B:=f^{-1}(A)$; we are to show that $B$ is closed. Let $(x_{n})_{n\in \mathbb {N} }$ be a sequence in $A$. By the characterisation of closedness by sequences in first-countable spaces, it will suffice to prove that $B$ is sequentially closed. Hence, let $(x_{n})_{n\in \mathbb {N} }$ be a sequence in $B$ that converges to some $x\in X$. Then $f(x_{n})\to f(x)$, so that $f(x)\in A$ and hence $x\in B$. $\Box$