The Set of Acc. Points under Homeomorphisms on Topological Spaces

# The Set of Accumulation Points under Homeomorphisms on Topological Spaces

Recall from the Homeomorphisms on Topological Spaces page that if $X$ and $Y$ are topological spaces then a bijective map $f : X \to Y$ is said to be a homeomorphism if it is continuous and open.

Furthermore, if such a homeomorphism exists then we say that $X$ and $Y$ are homeomorphic and write $X \simeq Y$.

We will now look at a nice topological property which says that if $f$ is a homeomorphism from $X$ to $Y$ and if $A$ is a subset of $X$ then the image of the set of accumulation points of $A$ is equal to the set of accumulation points of the image of $A$.

 Theorem 1: Let $X$ and $Y$ be topological spaces, let $f : X \to Y$ be a homeomorphism, and let $A \subseteq X$. Then $f(A') = (f(A))'$.
• Proof: Let $x \in f(A)'$. Then $f^{-1}(x) \in A'$, so $f^{-1}(x)$ is an accumulation point of $A$. So, for all open neighbourhoods $U$ in $X$ of $f^{-1}(x)$ we have that:
(1)
\begin{align} \quad A \cap U \setminus \{ f^{-1}(x) \} \neq \emptyset \end{align}
• Then we have that:
(2)
\begin{align} \quad f(A) \cap f(U) \setminus \{ x \} \neq \emptyset \end{align}
• Since $f$ is a homeomorphism and $U$ is an open set in $X$ we have that then $f(U)$ is an open set in $Y$. Furthermore, every open set in $Y$ is of the form $f(U)$ since $f$ is a bijection. Therefore $x$ is an accumulation point of $f(A)$, so $x \in (f(A))'$. Therefore:
(3)
\begin{align} \quad f(A') \subseteq (f(A))' \end{align}
• Now let $x \in (f(A))'$. Then $x$ is an accumulation point of $f(A)$ and so for every open neighbourhoods $V$ in $Y$ of $x$ we have that:
(4)
\begin{align} \quad f(A) \cap V \setminus \{ x \} \neq \emptyset \end{align}
• Therefore we have that:
(5)
\begin{align} \quad A \cap f^{-1} (V) \setminus \{ f^{-1} (x) \} \neq \emptyset \end{align}
• Since $f$ is a homeomorphism and $V$ is an open set we have that $f^{-1} (V)$ is an open set in $X$. Furthermore, every open set in $X$ is of this form since $f$ is a bijection. Therefore $f^{-1} (x)$ is an accumulation point of $A$, so $f^{-1}(x) \in A'$ and $x \in f(A')$. Therefore:
(6)
\begin{align} \quad (f(A))' \subseteq f(A') \end{align}
• We conclude that then $f(A') = (f(A))'$. $\blacksquare$