If $f$ is a continuous function from metric space $X$ to metric space $Y$, then $$f(\bar E)\subseteq \overline {f(E)}$$
for $E\subseteq X$.
To prove this, let $x \in f(\bar E)$, then $x=f(p)$ for $p\in E \cup E'$. If $p\in E$ then $x=f(p) \in \overline {f(E)}$.
If $p\in E'$ then $\forall \delta >0$ there is a neighborhood $N_{\delta}(p)=\{q : d(q,p) < \delta\}$ satisfying $$N_{\delta}(p) \cap E \neq \varnothing \Rightarrow f(N_{\delta}(p) \cap E)=f(N_{\delta}(p)) \cap f(E) \neq f(\varnothing)=\varnothing $$
Because $f$ is continuous, we know that $\forall \epsilon >0 \exists \delta > 0$ such that
$$f(N_{\delta}(p))\subseteq N_{\epsilon}(f(p))$$
which implies that
$$\varnothing \neq f(N_{\delta}(p))\cap f(E)\subseteq N_{\epsilon}(f(p)) \cap f(E) \neq \varnothing $$
therefore $x=f(p)\in f(E)'$ and we are done.
I like this way of expressing limit related arguments because it shortens proofs and reduces the amount of words required to explain reasoning. It also encapsulates and clarifies some definitions regarding limits and continuity, which allows for manipulation reminiscent of algebra.
However, I have no idea if these steps are justified. From what I have managed to find on the internet, the image and inverse image of the empty set is empty and $A\subseteq B \Rightarrow f(A)\subseteq f(B)$ so I believe my arguments are justified. If it is wrong, I would like to know why it is wrong so that I can maybe fix it and keep on using it for self-study purposes. If it is right, why is it not used more often?
If $f$ is a continuous function from metric space $X$ to metric space $Y$, then $f(\bar E)\subseteq \overline {f(E)}$ for $E\subseteq X$.
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real-analysis
proof-verification
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1Your proof is correct.It can be shortened.You do not need to distinguish the cases $x\in E$ and $x\in E'.$ If $x\in \bar E$ but $f(x)\not \in \bar f(E) $ then $f(x)$ has a nbhd $V$ with $ f(E)\cap V=\phi.$ ....... But if is continuous then $x$ has a nbhd $U$ with $f(U)\subset V.$ .......This results in a contradiction, because $x\in \bar E\implies \phi \ne U\cap E$ $\implies$ $ \phi \ne f(U\cap E)=$ $f(U\cap E)\cap V\subset f(E)\cap V,$ contrary to $\phi=F=f(E)\cap V$. – 2017-01-17
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1This is a property of all continuous functions on all spaces, and is equivalent to continuity. In general a function $f:X\to Y$ is defined to be continuous iff $f^{-1}V$ is open in $X$ whenever $V$ is open in $Y.$ Another equivalent is that $f^{-1}V$ is closed in $X$ whenever $V$ is closed in $Y.$ – 2017-01-17
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1In the second line of my first comment, $\bar f(E)$ should be $\overline {f(E)}. $ And the last formula should be $\phi=f(E)\cap V.$ – 2017-01-17