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$A$ is a subset of $\mathbb{R}$ with more than 1 point and $p: \mathbb{R} \to \mathbb{R}/A$ is the quotient map. Prove that $p$ is an open map $\iff$ $A$ is open.

I know if $A$ is open then for each open set $U$ of $\mathbb{R}$, if $U$ does not intersect $A$, we have $p^{-1}(p(U))=U$; if $U$ intersects $A$, we have $p^{-1}(p(U))=U\cup A$. Therefore p is an open map.

For the inverse direction, I tried if $A$ contains some open set $V$ of $\mathbb{R}$, then $p(V)$ is open so $p^{-1}(p(V))=A$ is open. If $A$ does not contain any open set, how can we say this case can not happen when $p$ is an open map?

Arthur
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user725757
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1 Answers1

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For the implication $p$ open $\implies$ $A$ open:

Let $a$ and $b$ be two distinct points of $A$ and let $U$ and $V$ be disjoint open neighbourhoods in ${\mathbb R}$ of, respectively, $a$ and $b$. The images $p(U)$ and $p(V)$ are, by assumption, open in ${\mathbb R}/A$. The intersection of these images is the one point set $\{A\}$, so this one point set is open as well. By definition of the quotient topology, this means that $A$ is open in ${\mathbb R}$.

Magdiragdag
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