testing for continuity via basic open sets
Proposition 1.
Let $X\mathrm{,}Y$ be topological spaces^{}, and $f\mathrm{:}X\mathrm{\to}Y$ a function. The following are equivalent^{}:

1.
$f$ is continuous^{};
 2.

3.
${f}^{1}(U)$ is open for any $U$ is a subbasis for the topology of $Y$.
Proof.
First, note that $(1)\Rightarrow (2)\Rightarrow (3)$, since every basic open set is open, and every element in a subbasis is in the basis it generates. We next prove $(3)\Rightarrow (2)\Rightarrow (1)$.

•
$(2)\Rightarrow (1)$. Suppose $\mathcal{B}$ is a basis for the topology of $Y$. Let $U$ be an open set in $Y$. Then $U$ is the union of elements in $\mathcal{B}$. In other words,
$$U=\bigcup \{{U}_{i}\in \mathcal{B}\mid i\in I\},$$ for some index set^{} $I$. So
${f}^{1}(U)$ $=$ ${f}^{1}({\displaystyle \bigcup \{{U}_{i}\in \mathcal{B}\mid i\in I\}})$ $=$ $\bigcup \{{f}^{1}({U}_{i})\mid i\in I\}}.$ By assumption^{}, each ${f}^{1}({U}_{i})$ is open, so is their union ${f}^{1}(U)$.

•
$(3)\Rightarrow (2)$. Suppose now that $\mathcal{S}$ is a subbasis, which generates the basis $\mathcal{B}$ for the topology of $Y$. If $U$ is a basic open set, then
$$U=\bigcap _{i=1}^{n}{U}_{i},$$ where each ${U}_{i}\in \mathcal{S}$. Then
${f}^{1}(U)$ $=$ ${f}^{1}({\displaystyle \bigcap _{i=1}^{n}}{U}_{i})$ $=$ $\bigcap _{i=1}^{n}}{f}^{1}({U}_{i}).$ By assumption, each ${f}^{1}({U}_{i})$ is open, so is their (finite) intersection^{} ${f}^{1}(U)$.
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Title  testing for continuity via basic open sets 

Canonical name  TestingForContinuityViaBasicOpenSets 
Date of creation  20130322 19:08:55 
Last modified on  20130322 19:08:55 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  4 
Author  CWoo (3771) 
Entry type  Result 
Classification  msc 26A15 
Classification  msc 54C05 