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

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 $ℬ$ is a basis for the topology of $Y$. Let $U$ be an open set in $Y$. Then $U$ is the union of elements in $ℬ$. In other words,

  $$U=\bigcup \{U_i\in ℬ\mid i\in I\},$$

  for some index set $I$. So

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

  $=$${\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 $𝒮$ is a subbasis, which generates the basis $ℬ$ for the topology of $Y$. If $U$ is a basic open set, then

  $$U=\bigcap _{i=1}^n{U}_i,$$

  where each $U_i\in 𝒮$. Then

  $f^{-1}(U)$$=$$f^{-1}({\displaystyle \bigcap _{i=1}^n}{U}_i)$

  $=$${\displaystyle \bigcap _{i=1}^n}{f}^{-1}(U_i).$

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

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