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This is for homework and I am wondering if my argument is correct.

Let $f :X \to Y$ be a continuous function. Show that $f({\overline{E}}) \subseteq \overline{f(E)}$ for any subset $E\subseteq X$. Give an example to show that the inclusion could be proper.\ If $E$ is closed then clearly the above is true. Instead suppose that $E$ is not closed. WTS for an arbitrary $y\in f(\overline{E})$, $y\in \overline{f(E)}$. Since $f$ is continuous $f(\overline{E})$ is closed. Suppose that $y\in f(\overline{E})^o$ (the interior of $f(\overline{E})$). Then $y\in f(E)$ and therefore also $y\in\overline{f(E)}$. Therefore $f(\overline{E}) \subseteq \overline{f(E)}$. Suppose that $y\in f(\overline{E})'$ (the set of limit points of $f(\overline{E})$). Then for some $x\in \overline{E}, f(x) = y$. This then means that for some $x\in E', f(x) = y\in f(E)'$ since $f$ is continuous. Therefore $y\in \overline{f(E)}$, and thus $f(\overline{E}) \subseteq \overline{f(E)}$.\

Any critiques would be greatly appreciated.

Jesse Jahn
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  • The image of a closed set under a continuous map is not necessarily closed. (The inverse image of a closed set under a continuous map is closed.) If $y \in f(\overline{E})$, then $y =f(x)$ for some $x \in \overline{E}$. To show $y \in \overline{f(E)}$, you need to show $y$ is the limit of a sequence in $f(E)$. How might the point $x\in \overline{E}$ help you do that? – kccu Dec 06 '15 at 22:47
  • You could also consider the complement $U$ of $\overline{f(E)}$ in $Y$. Since $f$ is continuous $V=f^{-1}(U)$ is open in $X$. – SamM Dec 06 '15 at 22:54
  • for $x\in \overline{E}$ that is a typo on my part, I will edit that and fix that. In response to the sequences, since f is continuous we know that for any element in X there is a sequence that converges to said point. Since we already know that f is continuous can we say that any element in Y has a sequence that converges to is using say for $x_{n}\to x$ then $f(x_{n})\to f(x)$? – Jesse Jahn Dec 06 '15 at 22:57
  • The $x\in\overline{E}$ is okay - see my earlier comment. And yes, since $f$ is continuous, if $x_n \to x$ then $f(x_n)\to f(x)$. (Since $x\in\overline{E}$, there exists a sequence ${x_n}$ in $E$ such that $x_n\to x$... Now use the continuity of $f$ as you suggested.) – kccu Dec 07 '15 at 02:34
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    Possible duplicate of Closure operation is not necessarily preserved –  Dec 13 '15 at 00:39

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