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Given homeomorphism $f: X \to Y$ and nowhere dense set $B \subset X$ show that $f(B)$ is nowhere dense set in $Y$.

I know that:

  • $f$ is homeomorphism $\implies$ $\left(U \text{ open in } X \iff f(U) \text{ open in } Y \right) (\star$)
  • homeomorphism preserves density ($\star \star$)

I tried using above two to prove it but it seems a little sketchy:

$$ B \text{ is nowhere dense in } X \implies \\ B^\complement \text{ is open and dense in } X \implies \\ f(B^\complement) \text{ is open and dense in } Y \text{ (using } \star \text { and } \star\star ) \implies \\ f(B) \text{ is nowhere dense in Y } $$

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  • $\begingroup$ @YuiToCheng is it really the proof? It seemed too simple for me so I expected a flaw in my reasoning. $\endgroup$ – math_beginner Apr 22 at 11:33
  • $\begingroup$ Notice a nowhere dense set needs not to be closed, e.g. $\{1/n\}_{n\in \Bbb N}$. See Henno Brandsma's answer for the correct definition. $\endgroup$ – YuiTo Cheng Apr 22 at 11:34
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    $\begingroup$ Your equivalence of nowhere dense is wrong: you need the complement of the closure of $B$ to be open and dense. $\endgroup$ – Henno Brandsma Apr 22 at 11:35
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    $\begingroup$ Use that $f[\overline{A}]=\overline{f[A]}$ and likewise for interior for all subsets $A$ of $X$. $\endgroup$ – Henno Brandsma Apr 22 at 11:37
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If $f: X \to Y$ is a homeomorphism and $B \subseteq X$ is nowhere dense then $f[B]$ is nowhere dense too:

$B$ is nowhere dense iff $\operatorname{int}(\overline{B}) = \emptyset$

Homeomorphisms preserve interiors and closures so

$\operatorname{int}(\overline{f[B]}) = \emptyset$ too and hence $f[B]$ is nowhere dense.

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  • $\begingroup$ $\operatorname{int}(\overline{f[B]}) = \operatorname{int}(f[\overline{B}])$ comes from another homeomorphism definition equivalence. How exactly did you conclude that $\operatorname{int}(\overline{B}) = \emptyset \implies \operatorname{int}(\overline{f[B]}) = \emptyset$ though? $\endgroup$ – math_beginner Apr 22 at 11:47
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    $\begingroup$ @Tomasz $f[\emptyset]=\emptyset$; apply $f$ to both sides of the definition of nowhere denseness of $B$ and apply the facts about interior, closures and $f$-images. $\endgroup$ – Henno Brandsma Apr 22 at 11:50
  • $\begingroup$ Thank you very much! $\endgroup$ – math_beginner Apr 22 at 11:51

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