Let $X$ be a Banach space, let $Y$ be a closed subspace in $X.$ Let a linear, continuous operator $\pi \colon X \to X/Y$ be defined by $\pi(x) = \overline{x}.$

I'd like to show $\pi$ open.

$\textbf{My attempt:}$

Take an open set $A \subset X$ and take $x \in A$. Therefore exists $\epsilon>0$ such that $B_{(x,\epsilon)} \subset A.$ Now I want to show that $\pi(B_{(x,\epsilon)}) = B_{(\overline{x},\epsilon)}$.

  • $\pi(B_{(x,\epsilon)}) \subset B_{(\overline{x},\epsilon)}:$

From continuity, I know the following inequality is true: $||\pi(x)|| \leq||x||$. Now take $y \in B_{(x, \epsilon)}$ and note that $$||\pi(x-y)||=||\pi(x)-\pi(y)|| \leq ||x-y|| < \epsilon.$$ Thefore $\pi(y) \in B_{(\overline{x},\epsilon)}$.

  • $B_{(\overline{x},\epsilon)} \subset \pi(B_{(x,\epsilon)}):$

Take $\overline{z} \in B_{(\overline{x},\epsilon)}$. I'm able to show that $\inf_{y \in Y} ||(z-x)+y||< \epsilon$, but can't progress any further.

Any suggestions on how to prove $(\supseteq)$?


Hint: We know that if $S \leq E$ is closed and $E$ is a Banach space, then the quotient $E/S$ is a Banach space as well with $\pi : E \to E/S$ a continuous linear mapping. In general, if you have any relation $\mathcal{R}$ on a set $X$, the mapping $x \in X \mapsto [x] \in X/\mathcal{R}$ is surjective. Now apply a famous theorem on surjective linear continuous mappings between Banach spaces.

  • $\begingroup$ I liked the solution, but do you think my path would not work? $\endgroup$ – Lucas Apr 1 at 19:07
  • 1
    $\begingroup$ I'd have to think about it, that's why I meant this as a hint. Note however that the inclusion you have proved says nothing about openness of $\pi$, in the sense that $\pi$ being open is equivalent to the other inclusion, maybe with a different $\varepsilon$. $\endgroup$ – Guido A. Apr 1 at 19:29

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