testing for continuity via nets
Proposition 1.
Let $X\mathrm{,}Y$ be topological spaces^{} and $f\mathrm{:}X\mathrm{\to}Y$. Then the following are equivalent^{}:

1.
$f$ is continuous^{};

2.
If $({x}_{i})$ is a net in $X$ converging to $x$, then $(f({x}_{i}))$ is a net in $Y$ converging to $f(x)$.

3.
Whenever two nets $({x}_{i})$ and $({y}_{j})$ in $X$ converge^{} to the same point, then $(f({x}_{i}))$ and $(f({y}_{j}))$ converge to the same point in $Y$.
Proof.
$(1)\iff (2)$. Let $A$ be the (directed) index set^{} for $i$. Suppose $f(x)\in U$ is open in $Y$. Then $x\in {f}^{1}(U)$ is open in $X$ since $f$ is continuous. By assumption^{}, $({x}_{i})$ is a net, so there is $b\in A$ such that ${x}_{j}\in {f}^{1}(U)$ for all $j\ge b$. This means that $f({x}_{j})\in U$ for all $i\ge b$, so $(f({x}_{i}))$ is a net too.
Conversely, suppose $f$ is not continuous, say, at a point $x\in X$. Then there is an open set $V$ containing $f(x)$ such that ${f}^{1}(V)$ does not contain any open set containing $x$. Let $A$ be the set of all open sets containing $x$. Then under reverse inclusion, $A$ is a directed set^{} (if ${U}_{1},{U}_{2}\in A$, then ${U}_{1}\cap {U}_{2}\in A$). Define a relation^{} $R\subseteq A\times X$ as follows:
$$(U,x)\in R\mathit{\hspace{1em}\hspace{1em}}\text{iff}\mathit{\hspace{1em}\hspace{1em}}x\in U{f}^{1}(V).$$ 
Then for each $U\in A$, there is an $x\in X$ such that $(U,x)\in R$, since $U\u2288{f}^{1}(V)$. By the axiom of choice^{}, we get a function $d\subseteq R$ from $A$ to $X$. Write $d(U):={x}_{U}$. Since $A$ is directed, $({x}_{U})$ is a net. In addition, $({x}_{U})$ converges to $x$ (just pick any $U\in A$, then for any $W\ge U$, we have $x\in W$ by the definition of $A$). However, $(f({x}_{U}))$ does not converge to $f(x)$, since ${x}_{U}\notin {f}^{1}(V)$ for any $U\in A$.
$(2)\iff (3)$. Suppose nets $({x}_{i})$ and $({y}_{j})$ both converge to $z\in X$. Then, by assumption, $(f({x}_{i}))$ and $(f({y}_{j}))$ are nets converging to $f(z)\in Y$.
Conversely, suppose $({x}_{i})$ converges to $x$, and $i$ is indexed by a directed set $A$. Define a net $({y}_{i})$ such that ${y}_{i}=x$ for all $i\in A$. Then $({y}_{i})=(x)$ clearly converges to $x$. Hence both $(f({x}_{i}))$ and $(f({y}_{i}))$ converge to the same point in $Y$. But $(f({y}_{i}))=(f(x))$ converges to $f(x)$, we see that $(f({x}_{i}))$ converges to $f(x)$ as well. ∎
Remark. In particular, if $X,Y$ are first countable, we may replace nets by sequences in the proposition^{}. In other words, $f$ is continuous iff it preserves converging sequences.
Title  testing for continuity via nets 

Canonical name  TestingForContinuityViaNets 
Date of creation  20130322 19:08:58 
Last modified on  20130322 19:08:58 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  7 
Author  CWoo (3771) 
Entry type  Result 
Classification  msc 54C05 
Classification  msc 26A15 
Related topic  Net 
Related topic  FirstCountable 