PROBLEM: Given a function $f:A\rightarrow N$, if there exist $\lim_{x\to a} f(x)=b$, then $b\in \overline{f(A)}$
Im not really sure how to proceed, I was thinking that I could split in cases, if $a\in A$ and if $a\in \overline{A}-A$. Then im pretty sure that I must use the fact that $f(A)\subset f(\overline{A})$ and if the function is continuous, $f(\overline{A})\subset \overline{f(A)}$ but I cant manage to join both things. I suppose its an easy question, hope you guys can help me.
Thanks in advance :)
A topological definition of $\lim_{x\rightarrow a} f(x) = b$:
For all open neighborhoods $V$ of $b$, there exists some open neighborhood $U$ of $a$ with $f(U - \{a\})$ as a subset of $V$.
Assume $a$ is not isolated, a necessary condition to prove the proposition.
So if $V$ is an open neighborhood of $b$, there's an
open $U$ neighborhood of $a$ with $f(U - \{a\})$ as a subset of $V$. Since $a$ is not isolated, $f(U - {a})$ is not empty.
Let $x$ be any point in $f(U - \{a\})$.
Thus $x$ is in $V \cap f(A)$. Conclude $b$ is in the closure of $f(A)$.
No. You have not been given that $f$ is continuous.
Use of sequences is a failure unless the spaces are first countable.