Limit point is a topological invariant, am I right?

Question

My proof: $$\text{Let }(X,T)\cong (Y,T^*).$$ $$\text{Let }H\in T,\,p\in H, A\subseteq X, p\in A. f:(X,T)\to(Y,T^*) \text{ a homeomorphism.}\\ \text{Then } H\cap[A\setminus\{p\}]\neq\emptyset$$ Then I prooved since f is bijective: $$f(H\cap[A\setminus\{p\}]) = f(H)\cap[f(A)\setminus f(\{p\})]\neq\emptyset$$ If I put $G\in T^*$ since $H$ is an arbitrary open subset of $X$, because $f$ is bijective$f^{-1}(G)=H_G$ for some $H_G$ like the $H$ I defined above. Therefore $$G\cap[f(A)\setminus\{f(p)\}]\neq\emptyset$$ So $f(p)$ is a limit point of $f(A)$ in $(Y,T^*)$ and that proves that limit point is a topological property. Am I right in my proof?

Answer 1

I cannot say that your proof is false. But it does not mean that I can give you a good grade for the proof. Compare your proof with the following:

Question. Let $f\colon X\to Y$ be a homeomorphism and let $A\subset X$. If $p$ is a limit point of $A$ in $X$, show that $f(p)$ is a limit point of $f(A)$ in $Y$.

Answer. For any open set $V$ in $Y$ containing $f(p)$, we need to show that $$ V\cap \bigl[ f(A)\setminus\{f(p)\} \bigr] \neq \varnothing $$ Since $f$ is bijective, we have $$ \begin{align*} f^{-1}\bigl(V\cap \bigl[ f(A)\setminus\{f(p)\} \bigr]\bigr) &= f^{-1}(V)\cap f^{-1}\bigl[ f(A)\setminus\{f(p)\} \bigr] \\ &= f^{-1}(V)\cap \bigl[ f^{-1}f(A) \setminus f^{-1}f(p) \bigr] \\ &= f^{-1}(V)\cap \bigl[ A \setminus \{p\} \bigr] \end{align*} $$ Moreover, $f^{-1}(V)$ is open in $X$ since $f$ is continuous, and $p$ is a limit point of $A$ in $X$ by assumption. Thus $$ f^{-1}\bigl(V\cap \bigl[ f(A)\setminus\{f(p)\} \bigr]\bigr) = f^{-1}(V)\cap \bigl[ A \setminus \{p\} \bigr] \neq \varnothing $$ and it implies that $V\cap \bigl[ f(A)\setminus\{f(p)\} \bigr] \neq \varnothing$ which we wanted to show.