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Can someone check my proof?

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $$f(\overline{E}) \subset \overline{f(E)} $$ for every set $E\subset X$. ($\overline{E}$ denotes the closure of $E$.)

Proof: Suppose $x\in \overline{E} = E\cup E'$, where $E'$ is the set of limit points of $E$.

If $x\in E$, then $f(x)\in f(E)\subset \overline{f(E)}$ and we are done.

Now suppose $x\in E'$ and let $\epsilon >0$ be given. Since $f$ is continuous, there exists a $\delta >0$ such that $d(x,y)<\delta$ implies $$d(f(x),f(y))<\epsilon. $$ Since $x$ is a limit point of $E$, there is a neighborhood $N_{\delta}(x)$ of radius $\delta$ about $x$ such that $(N_{\delta}(x)\setminus \{x\}) \cap E \neq \emptyset$. Then $p\in N_{\delta}(x)$ implies that $d(f(x),f(p))<\epsilon$.

Hence, for any $\epsilon >0$, we can always find a $\delta$ such that $$\left( N_{\epsilon}(f(x)) \setminus \{f(x)\}\right) \cap f(E) \neq \emptyset.$$ Thus $f(x)\in f(E)'\subset \overline{f(E)}$.

I'm not really sure about the last part, how can we make certain that if $p\in N_{\delta}(x) \setminus \{x\}$, then $f(p) \in N_{\epsilon}(f(x))\setminus \{f(x)\}$. Since we aren't given that $f$ is injective, we can have $x\neq p$ and $f(x) = f(p)$? Any input or alternative approaches would be greatly appreciated!

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  • $\begingroup$ You should reword the sentence that starts with "Since $x$ is a limit point of $E$", because to be a limit point of $E$, every neighborhood around $x$ intersects $E$ at a point other than $x$. You made it seem like $x$ being a limit point means only that there exists a neighborhood with this property -- actually, it means something stronger: that every neighborhood around $x$ has this property. $\endgroup$
    – layman
    May 14, 2015 at 2:58
  • $\begingroup$ Maybe re-word it to: Since $x$ is a limit point of $E$, and $N_{\delta}(x)$ is a $\delta$-neighborhood of $x$, we know $(N_{\delta}(x) - \{ x \} ) \cap E \neq \emptyset$. $\endgroup$
    – layman
    May 14, 2015 at 3:00
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    $\begingroup$ Yes, it is possible that $f(x) = f(p)$, so you can only conclude that $N_{\epsilon}(f(x))\cap f(E) \neq \emptyset$, which means that $f(x) \in \overline{f(E)}$, not necessarily in $f(E)'$. $\endgroup$ May 14, 2015 at 3:06

3 Answers 3

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Since $f(E)\subset \overline{f(E)}$ then, $$E\subset f^{-1}(f(E))\subset f^{-1}(\overline{f(E)}).$$ Since $f$ is continuous and $\overline{f(E)}$ is closed, then $f^{-1}(\overline{f(E)})$ is closed. Thus, $\overline{E}\subset f^{-1}(\overline{f(E)})$ and therefore $f(\overline{E})\subset \overline{f(E)}$.

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if $x_n \to x$ then $f(x_n) \to f(x).$ This is essentially what continuity means.

So if $x_n \in E, \forall n$, then $f(x_n) \in f(E)$ and $f(x) \in \overline{f(E)}.$

Since any $x$ in $\bar{E}$ is the limit of such a sequence, we are done.

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Let $y \in f(\overline{E}) \Rightarrow y = f(x), x \in \overline{E}$. If $x \in E \to y=f(x) \in f(E) \subset \overline{f(E)}$. If $x \in E'$, then there is a sequence $\{x_n\} \in E$ such that $x_n \to x \to f(x_n)\to f(x)$ since $f$ is continuous. Thus $y = f(x)$ is a limit point of $f(E) \to y \in f(E)' \subset \overline{f(E)}$. Thus $f(\overline{E}) \subset \overline{f(E)}$.

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