# Equivalence of Definitions of Homeomorphic Metric Spaces

## Theorem

The following definitions of the concept of Homeomorphic Metric Spaces are equivalent:

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f: A_1 \to A_2$ be a bijection.

### Definition 1

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f: A_1 \to A_2$ be a bijection such that:

$f$ is continuous from $M_1$ to $M_2$
$f^{-1}$ is continuous from $M_2$ to $M_1$.

Then:

$f$ is a homeomorphism
$M_1$ and $M_2$ are homeomorphic.

### Definition 2

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f: A_1 \to A_2$ be a bijection such that:

for all $U \subseteq A_1$, $U$ is an open set of $M_1$ if and only if $f \left[{U}\right]$ is an open set of $M_2$.

Then:

$f$ is a homeomorphism
$M_1$ and $M_2$ are homeomorphic.

### Definition 3

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f: A_1 \to A_2$ be a bijection such that:

for all $V \subseteq A_1$, $V$ is a closed set of $M_1$ if and only if $f \left[{V}\right]$ is a closed set of $M_2$.

Then:

$f$ is a homeomorphism
$M_1$ and $M_2$ are homeomorphic.

### Definition 4

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f: A_1 \to A_2$ be a bijection such that:

for all $a \in A_1$ and $N \subseteq A_1$, $N$ is a neighborhood of $a$ if and only if $f \left[{N}\right]$ is a neighborhood of $f \left({a}\right)$.

Then:

$f$ is a homeomorphism
$M_1$ and $M_2$ are homeomorphic.

## Proof

In order to prove the assertion it is sufficient to prove that the conditions for homeomorphism in definitions $2$ to $4$ are necessary and sufficient conditions for $f$ and $f^{-1}$ to be continuous on $M_1$ and $M_2$ respectively.

### Definition 1 implies Definition 2

Let $f$ and $f^{-1}$ both be continuous by definition $2$.

Let $U \subseteq A_1$ be open in $M_1$.

As $f^{-1}$ is continuous, $\left({f^{-1}}\right)^{-1} \left[{U}\right] = f \left[{U}\right]$ is open in $M_2$.

That is, $f \left[{U}\right] = f \left[{U}\right]$ is open in $M_2$.

Let $f \left[{U}\right] \subseteq A_2$ be open in $M_2$.

Then as $f$ is continuous, $f^{-1} \left[{f \left[{U}\right]}\right] = U$ is open in $M_1$.

Thus:

for all $U \subseteq A_1$, $U$ is an open set of $M_1$ if and only if $f \left[{U}\right]$ is an open set of $M_2$.

$\Box$

### Definition 2 implies Definition 4

Let definition $2$ hold.

Let $a \in A_1$.

Let $N \subseteq A_1$.

Then:

$N$ is a neighborhood of $a$
$N$ contains an open set $U$ containing $a$
$f \left[{N}\right]$ contains an open set $U' = f \left[{U}\right]$ containing $f \left({a}\right)$
$f \left[{N}\right]$ is a neighborhood $f \left({a}\right)$.

$\Box$

### Definition 4 implies Definition 1

Let definition $4$ hold.

Let $a \in A_1$.

Let $U \subseteq A_2$ be a neighborhood of $f \left({a}\right)$.

Then $f \left[{f^{-1} \left[{U}\right]}\right]$ is a neighborhood $f \left({a}\right)$.

Hence $f^{-1} \left[{U}\right]$ is a neighborhood of $a$.

Thus $f$ is continuous.

Similarly, let $b \in A_2$.

Let $V \subseteq A_1$ be a neighborhood of $f^{-1} \left({b}\right)$.

Then $f^{-1} \left[{f \left[{V}\right]}\right]$ is a neighborhood $f^{-1} \left({b}\right)$.

Hence $f \left[{V}\right]$ is a neighborhood of $f \left({f^{-1} \left({b}\right)}\right) = b$.

Thus $f^{-1}$ is continuous.

$\Box$

### Definition 2 iff Definition 3

This is demonstrated in Continuity of Mapping between Metric Spaces by Closed Sets.

$\blacksquare$