Question about proof that $f$ continuous $\implies f(\overline{A})\subset \overline{f(A)}$ $f$ continuous $\implies f(\overline{A})\subset \overline{f(A)}$ for $A\subset X$
My book presents the following proof:
Supposing $f$ continuous, and suppose $V$ an open containing $f(V)$. Then, $f^{-1}(V)$ is an open in $x$ that contains $x$. Therefore, it must intersect $A$ in some point $y$. Then, $V$ intersect $f(A)$ at the point $f(y)$, therefore $f(x)\in \overline{f(A)}$.
First of all, in the second line, "it must intersect $A$ in some point$. Why?
Also, why "then, $V$ intersect $f(A)$ at some point $f(y)$"?
I understand that the proof goes by saying that if $x\in \overline{A}$ then $f(x)\in \overline{f(A)}$. I just didn't understand the steps mentioned.
 A: Clearly, $A \subset f^{-1}(f(A)) \subset f^{-1}(\overline{f(A)})$. Since $f$ is continuous, $f^{-1}(\overline{f(A)})$ is closed and hence $\overline{A} \subset f^{-1}(\overline{f(A)})$. It follows that $f(\overline{A})\subset f(f^{-1}(\overline{f(A)})) \subset \overline{f(A)}$
A: See, to show that $f(x) \in \overline{f(A)}$, we have to show that every open set contaning $f(x)$ intersects $f(A)$.
To do this, we take such a neighbourhood, called $V$, which contains $f(x)$. (The question has a misprint) .
Now, because $V$ is open and $f$ is continuous, $f$ takes open sets back to open sets (I think you can prove this yourself. If you can't then reply back).
Which means that $f^{-1}(V)$ is open in $X$.  
All we need now is that because $f^{-1}(V)$ is an open neighbourhood of $x$, and $x \in \overline{A}$, the set $f^{-1}(V) \cap A$ is non-empty, hence $\exists y \in f^{-1}(V)$ ( it is possible that $ y = x$.)
Now, $y \in f^{-1}(V)$ means that $f(y) \in V$, hence $V$ intersects $A$ at the point $f(y)$.
A: The proof uses that the following statements are equivalent:


*

*$x \in \overline A$

*For every open set $U$, if $x \in U$ then $U \cap A \neq \varnothing$.


There is an unwritten assumption at the start of the proof that $x \in \overline A$.
For your first question, since $f^{-1}(V)$ is an open set containing $x$ and $x \in \overline A$, it follows that $f^{-1}(V)$ must intersect $A$ at some point $y$ because $f^{-1}(V) \cap A \neq \varnothing$.
For your second question, note that it is not at some point $f(y)$, but rather at the point $f(y)$ with $y$ being the same as in the previous step.
You just need to show that if $y \in f^{-1}(V) \cap A$ then $f(y) \in V \cap f(A)$, which follows immediately from the definitions of $f^{-1}(V)$ and $f(A)$.
A: Is that exactly what the book says? If so, it needs a total re-write. 
Let $x\in \bar A.$ Let $V$ be a nbhd of $f(x).$ Then $f^{-1}(V)$ is a nbhd of $x,$ and $x\in \bar A,$ so $A\cap f^{-1}(V)\ne \phi.$  So $$\phi \ne f(A\cap f^{-1}V)\subset f(A)\cap f(f^{-1}(V))\subset f(A)\cap V.$$ So every nbhd $V$ of $f(x)$ intersects $f(A).$ That is,  $f(x)\in \overline {f(A)}. $ This holds for every  $x\in \bar A, $ so $f(\bar A)\subset \overline {f(A)}.$
Remarks: (1). This applies for any continuous function $f:X\to Y.$ We just amend it  to say $x$ is in the closure of $A$ in the space $X,$ that $V$ is a nbhd of $f(x)$ in the space $Y$, that $f^{-1}(V)$ is a nbhd of $x$ in the space $X,$ and that $\overline {f(A)}$ means the closure of $f(A)$ in $Y.$
(2).The converse also holds: For any $f:X\to Y,$ if $f(\bar A)\subset \overline {f(A)}$ for every $A\subset X,$ then $f$ is continuous.
