The Closure of a Set under Homeomorphisms on Topological Spaces

# The Closure of a Set 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 closure of $A$ is equal to the closure 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 $f(\bar{A}) = \overline{f(A)}$.
• Proof Let $x \in f(\bar{A})$. Then $f^{-1}(x) \in \bar{A}$, i.e., $f^{-1}(x)$ is in the closure of $A$. So $f^{-1}(x) \in A \cup A’$. If $f^{-1}(x) \in A$ then $x \in f(A)$ and $f(A) \subseteq \overline{f(A)}$, so $x \in \overline{f(A)}$. If $f^{-1}(x) \in A’$ then $f^{-1}(x)$ is an accumulation point of $A$ so for every open set $U$ of $X$ with $f^{-1}(x) \in U$ we have that:
(1)
\begin{align} \quad A \cap U \setminus \{ f^{-1}(x) \} \neq \emptyset \end{align}
• So we have that:
(2)
\begin{align} \quad f(A \cap U \setminus \{ f^{-1}(x) \}) = f(A) \cap f(U) \setminus \{ x \} \neq \emptyset \end{align}
• Since $f$ is a homeomorphism we have that since $U$ is an open set in $X$ that then $f(U)$ is an open set in $Y$. But every open set on $Y$ is of the form $f(U)$ since $f$ is a bijection. Therefore $x$ is an accumulation point of $f(A)$ so $x \in (f(A))’$. But $(f(A))’ \subseteq f(A) \cup (f(A))’ = \overline{f(A)}$. Hence $x \in \overline{f(A)}$ which shows that:
(3)
\begin{align} \quad f(\bar{A}) \subseteq \overline{f(A)} \end{align}
• Now let $x \in \overline{f(A)}$. Then $x$ is an accumulation point of $f(A)$, so $x \in f(A) \cup (f(A))’$. If $x \in f(A)$ then $f^{-1}(x) \in A$. But $A \subseteq A \cup A’ = \bar{A}$ so $f^{-1}(x) \in \bar{A}$ and $x \in f(\bar{A})$. If $x \in (f(A))’$ then $x$ is an accumulation point of $f(A)$ and so for every open set $V$ of $Y$ with $x \in V$ we have that:
(4)
\begin{align} \quad f(A) \cap V \setminus \{ x \} \neq \emptyset \end{align}
• So then:
(5)
\begin{align} A \cap f^{-1}(V) \setminus \{ f^{-1}(x) \} \neq \emptyset \end{align}
• Since $f$ is a homeomorphism and $V$ is an open set in $Y$ we have that $f^{-1}(V)$ is an open set in $X$. But every open set in $X$ is of the form $f^{-1}(V)$ since $f$ is a bijection. Therefore $f^{-1}(x)$ is an accumulation point of $A$ so $f^{-1} (x) \in A’$. Hence $x \in f(A’)$. But $f(A’) \subseteq f(A \cup A’) = f(\bar{A})$. Hence $x \in f(\bar{A})$ and so:
(6)
\begin{align} \quad \overline{f(A)} \subseteq f(\bar{A}) \end{align}
• We therefore conclude that $f(\bar{A}) = \overline{f(A)}$. $\blacksquare$