[증명] The metric function is continuous

Proposition

Let $X$ be a metric space with the metric $d$.

$d : X \times X \rightarrow \mathbb{R}$ is continuous.

 

 

Proof

The basis for $\mathbb{R}$ is $(a, b)$. If $\forall a, b$, $d^{-1}((a, b))$ is open in $X \times X$, then d is continuous.

 

Case 1. $b<0$.

$d^{-1}((a, b))$ is open (empty set).

 

 

Case 2. $a < 0 \leq b$

$\forall x \times y \in d^{-1}((a, b))$, $0 \leq d(x, y) < b$

Also, $\forall x_1 \in B(x, \frac{b-d(x, y)}{2}), y_1 \in B(y, \frac{b-d(x, y)}{2})$, 

$d(x_1, y_2) < d(x_1, x) + d(x, y) + d(y, y_1)$ (triangular inequality)

Here, $d(x_1, x) < \frac{b-d(x, y)}{2}$ and $d(y, y_1) < \frac{b-d(x, y)}{2} \Rightarrow d(x_1, y_1) < b$.

$\Rightarrow \exists$ open set $B(x, \frac{b-d(x, y)}{2}) \times B(y, \frac{b-d(x, y)}{2}) \subset d^{-1}((a, b))$

$\Rightarrow d^{-1}((a, b))$ is open.

 

 

Case 3. $0 \leq a < b$

$\forall x \times y \in d^{-1}((a, b))$, $a < d(x, y) < b$

Also, $\forall x_1 \in B(x, \epsilon), y_1 \in B(y, \epsilon)$, 

Where $\epsilon = min( \frac{b-d(x, y)}{2}, \frac{d(x, y)-a}{2} )$

 

$d(x_1, y_2) < d(x_1, x) + d(x, y) + d(y, y_1)$ (triangular inequality)

Here, $d(x_1, x) < \frac{b-d(x, y)}{2}$ and $d(y, y_1) < \frac{b-d(x, y)}{2} \Rightarrow d(x_1, y_1) < b$.

 

$d(x, y) < d(x, x_1) + d(x_1, y_1) + d(y_1, y)$ (triangular inequality)

Here, $d(x_1, x) < \frac{d(x, y)-a}{2}$ and $d(y, y_1) < \frac{d(x, y)-a}{2} \Rightarrow a < d(x_1, y_1)$.

 

$\Rightarrow \exists$ open set $ B(x, \epsilon) \times B(y, \epsilon) \subset d^{-1}((a, b))$

$\Rightarrow d^{-1}((a, b))$ is open.

 

 

$\Rightarrow d$ is continuous.

 

증명에 문제가 있다면 댓글로 의견 부탁드립니다.

Please let me know any errors in the proof in the comments.