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Hometesting for continuity via basic open sets
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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. $f$ 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 $\mathcal{B}$ 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 $f^{{1}}(U)$.
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