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Hometesting for continuity via nets

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# testing for continuity via nets

###### Proposition 1.

Let $X,Y$ be topological spaces and $f:X\to Y$. 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.

$(1)\Leftrightarrow(2)$. 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 $j\geq b$. This means that $f(x_{j})\in U$ for all $i\geq b$, 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:

$(U,x)\in R\qquad\mbox{iff}\qquad x\in U-f^{{-1}}(V).$ |

Then for each $U\in A$, there is an $x\in X$ such that $(U,x)\in R$, since $U\not\subseteq f^{{-1}}(V)$. 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 $W\geq U$, 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 $U\in A$.

$(2)\Leftrightarrow(3)$. 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.

## Mathematics Subject Classification

54C05*no label found*26A15

*no label found*

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