testing for continuity via basic open sets

Proposition 1.

Let $X,Y$ be topological spaces, and $f:X\to Y$ a function. The following are equivalent:

1. 1.

$f$ is continuous;

2. 2.

$f^{-1}(U)$ is open for any $U$ in a basis ($U$ called a basic open set) for the topology of $Y$;

3. 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

 $\displaystyle f^{-1}(U)$ $\displaystyle=$ $\displaystyle f^{-1}(\bigcup\{U_{i}\in\mathcal{B}\mid i\in I\})$ $\displaystyle=$ $\displaystyle\bigcup\{f^{-1}(U_{i})\mid i\in I\}.$

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

 $\displaystyle f^{-1}(U)$ $\displaystyle=$ $\displaystyle f^{-1}(\bigcap_{i=1}^{n}U_{i})$ $\displaystyle=$ $\displaystyle\bigcap_{i=1}^{n}f^{-1}(U_{i}).$

By assumption, each $f^{-1}(U_{i})$ is open, so is their (finite) intersection $f^{-1}(U)$.

Title testing for continuity via basic open sets TestingForContinuityViaBasicOpenSets 2013-03-22 19:08:55 2013-03-22 19:08:55 CWoo (3771) CWoo (3771) 4 CWoo (3771) Result msc 26A15 msc 54C05