# The Interior 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 begin by looking 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 interior of $A$ is equal to the interior 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 $f(\mathrm{int}(A)) = \mathrm{int}(f(A))$. |

**Proof:**Let $x \in f(\mathrm{int}(A))$. Then $f^{-1}(x) \in \mathrm{int}(A)$ so $f^{-1}(x)$ is an interior point of $A$ and so there exists an open neighbourhood $U$ in $X$ of $f^{-1}(x)$ such that:

(1)

\begin{align} \quad f^{-1}(x) \in U \subseteq A \end{align}

- Hence we have that $x \in f(U) \subseteq f(A)$. Since $f$ is a homeomorphism and $U$ is open in $X$ we have that $f(U)$ is open in $Y$ so $f(U)$ is an open neighbourhood of $x$ contained in $f(A)$. Hence $x$ is an interior point of $f(A)$ so $x \in \mathrm{int} (f(A))$ so:

(2)

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

- Now let $x \in \mathrm{int} (f(A))$. Then $x$ is an interior point of $f(A)$ and so there exists an open neighbourhood $V$ in $Y$ of $x$ such that:

(3)

\begin{align} \quad x \in V \subseteq f(A) \end{align}

- Therefore $f^{-1}(x) \in f^{-1}(V) \subseteq A$. Since $f$ is a homeomorphism and $V$ is open in $Y$ we have that $f^{-1}(V)$ is open in $X$. Therefore $f^{-1}(V)$ is an open neighbourhood of $f^{-1}(x)$ contained in $A$ and therefore $f^{-1}(x) \in \mathrm{int}(A)$ so $x \in f(\mathrm{int}(A))$. Therefore:

(4)

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

- We hence conclude that $f(\mathrm{int}(A)) = \mathrm{int} (f(A))$. $\blacksquare$