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$f: X \rightarrow Y$ is continuous between two topological spaces. Show $\forall A \subseteq X$ we have $f(\overline{A}) \subseteq \overline{f(A)}$

Since we know that the function is continuous, we know that the preimage of a closed set in $Y$ is a closed set in $X$.

However, I don't see how this helps.

Another idea I have is to show that if every open set $V \subseteq Y$ containing $x$ where $x \in f(\overline{A})$ we have: $V \cup f(A) \ne \emptyset$. That would imply that $x \in \overline{f(A)}$

However, I'm not quite sure if this is the most intuitive apprach.

Where would I start to show this inclusion?

ViktorStein
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HoopsMcCann
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5 Answers5

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You're on the right path but it needs more details in this implication you made (assuming that there is a typo in your question, ie, you wanted to write $f(A) \cap V \neq \emptyset$). What you need to show is: Given $y \in f(\overline{A})$, for every open set $V \subset Y$ such that $y \in V$ you have $V \cap f(A) \neq \emptyset$. This is true since $y \in f(\overline{A})$ implies that $y = f(x)$ with $x \in \overline{A}$, as $f^{-1}(V)$ is an open set (f is a continuous function) and $x \in f^{-1}(V)$ there exists $x_{1} \in f^{-1}(V) \cap A$ which gives you that $f(x_{1}) \in V \cap f(A)$, ie, $V \cap f(A) \neq \emptyset$.

ABP
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  1. $f[A] \subseteq \overline{f[A]}$ (definition of closure)

  2. $A \subseteq f^{-1}[f[A]]$ (by definition of $f^{-1}$)

  3. Combining 1 and 2 $A \subseteq f^{-1}[\overline{f[A]}]$.

  4. $f^{-1}[\overline{f[A]}]$ is closed as $\overline{f[A]}$ is closed and $f$ is continuous.

  5. $\overline{A}$ is the smallest closed set containing $A$ so $$\overline{A} \subseteq f^{-1}[\overline{f[A]}]$$

  6. From 5. by definition of $f^{-1}$ we have $f[\overline{A}] \subseteq \overline{f[A]}$.

QED.

Or if you know theory about nets: if $x \in \overline{A}$ there is some net $(a_i)_i$ from $A$ such that $a_i \to x$. Continuity of $f$ preserves net convergence so $f(a_i) \to f(x)$ and as all $f(a_i)$ lie in $f[A]$ by definition, we conclude $f(x) \in \overline{f[A]}$. The inclusion has been shown.

Henno Brandsma
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HINT:

$f(A)\subset \overline {f(A)}$, so $A \subset f^{-1}(\overline {f(A)})$. But $f^{-1}(\overline {f(A)})$ is closed, so $\overline A \subset f^{-1}(\overline {f(A)})$, and so $f(\overline A) \subset \overline {f(A)}$.

orangeskid
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If f is continuous, then for $A\subseteq X$, $\overline{f(A)}$ is closed in Y. Hence $f^{-1}(\overline{f(A)})$ is closed in X and contains $\overline{A}$

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Let $f(B)=\overline{f(A)}$, then by continuity B is closed and $A\subseteq B$. As $\overline{A}$ is an intersection of all closed sets containing $A$, we get $\overline{A}\subseteq B$ and $f(\overline{A})\subseteq f(B)=\overline{f(A)}$

Wisielec
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  • How do you justify the existence of this set $B$ such that $\overline{f(A)} = f(B)$? – ABP Feb 24 '20 at 00:01
  • It is the preimage of $\overline{f(A)}$ – Wisielec Feb 24 '20 at 00:08
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    Take $f:\mathbb{R} \to\mathbb{R}$ given by $f(x) =c$ for all $x\in\mathbb{R}$, put $A=(0, 2)$ and $B=(0, 1)$. We have $f$ contiuous and $f(B) =\overline{f(A)}$. But is it $B$ closed or is it $A\subset B$? – freshman Feb 24 '20 at 00:13
  • You don't have that $f(f^{-1}(D)) = D$ in general (you're applying this with $D = \overline{f^{-1}(A)}$). In order to have this you'd need that your function is bijective, what you don't have and don't need to prove the result in this question. – ABP Feb 24 '20 at 00:18
  • By $B$ I meant the preimage of $\overline{f(A)}$, then for $f(x)=c$ we get $B=\mathbb{R}$, so $B$ is closed and contains $A$. But I can't prove that this preimage exists in general case. – Wisielec Feb 24 '20 at 00:33