The Boundary of a Set under Homeomorphisms on Topological Spaces

# The Boundary 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 boundary of $A$ is equal to the boundary 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 $f(\partial A) = \partial(f(A))$. |

*We will prove Theorem 1 in two different ways.*

**Proof 1):**Let $x \in f(\partial A)$. Then $f^{-1}(x) \in \partial A$. So $f^{-1}(x)$ is a boundary point of $A$, so $f^{-1}(x) \in \bar{A} \setminus \mathrm{int}(A)$. So:

\begin{align} \quad x \in f(\bar{A}) \setminus f(\mathrm{int}(A)) \end{align}

- By the theorems presents on The Closure of a Set under Homeomorphisms on Topological Spaces and The Interior of a Set under Homeomorphisms on Topological Spaces pages, since $f$ is a Homeomorphism we have that:

\begin{align} \quad f(\bar{A}) = \overline{f(A)} \quad \mathrm{and} \quad f(\mathrm{int}(A)) = \mathrm{int}(f(A)) \end{align}

- Hence we get that $x \in \overline{f(A)} \setminus \mathrm{int}(f(A)) = \partial (f(A))$ and so:

\begin{align} \quad f(\partial A) \subseteq \partial (f(A)) \end{align}

- Now let $x \in \partial (f(A))$. Then $x$ is a boundary point of $f(A)$ so:

\begin{align} \quad x \in \overline{f(A)} \setminus \mathrm{int}(f(A)) \end{align}

- By the theorems referenced above, since $f$ is a homeomorphism we have that $\overline{f(A)} = f(\bar{A})$ and $\mathrm{int}(f(A)) = f(\mathrm{int} (A))$. Therefore $x \in f(\bar{A}) \setminus f(\mathrm{int}(A))$ so $f^{-1}(x) \in \bar{A} \setminus \mathrm{int} (A) = \partial A$. So $x \in f(\partial A)$, and hence:

\begin{align} \quad \partial (f(A)) \subseteq f(\partial A) \end{align}

- We conclude that $f(\partial A) = \partial (f(A))$. $\blacksquare$

**Proof 2)**Let $x \in \partial A$. Then $f(x) \in f(\partial A)$. Let $V$ be any open neighbourhood of $f(x)$ in $Y$. Since $f$ is continuous, $f^{-1}(V)$ is open in $X$ and $x \in f^{-1}(V)$. So there exists points $a, b \in X$ with $a \in A \cap f^{-1}(V)$ and $b \in A^c \cap f^{-1}(V)$ where $a, b \neq x$ since $f$ is bijective. Therefore $f(a) \in f(A) \cap U$ and $f(b) \in (f(A))^c \cap U$ where $f(a), f(b) \neq f(x)$. So $f(x) \in \partial (f(A))$ which shows that:

\begin{align} \quad f(\partial A) \subseteq \partial (f(A)) \end{align}

- Now let $x \in \partial (f(A))$ and let $V$ be an open neighbourhood of $Y$. Then there exists $a, b \in Y$ with $a \in f(A) \cap V$ and $b \in f(A^c) \cap V$ where $a, b \neq x$. Then $f^{-1}(a) \in A \cap f^{-1}(V)$ and $f^{-1}(b) \in A^c \cap f^{-1}(V)$ and since $f$ is continuous, $f^{-1}(V)$ is open in $X$ which shows that $f^{-1}(x) \in \partial A$. So $x \in f(\partial A)$ which shows that:

\begin{align} \partial (f(A)) \subseteq f(\partial A) \end{align}

- So we conclude that $f(\partial A) = \partial (f(A))$. $\blacksquare$