Criticize my proof (metric topology, Munkres)

[mathmonkey]

Homework Statement

Let $X$ be a metric space with metric $d$. Show that $d: X \times X \mapsto \mathbb{R}$ is continuous.

Homework Equations

The Attempt at a Solution

Please try to poke holes in my proof, and if it is correct, please let me know if there's any more efficient way to do it. Any help would be greatly appreciated! Here is my proof:

First, we show that for any fixed $\textbf{x}$, the set $A(\textbf{x},\epsilon)$ defined as $\{\textbf{y}|d(\textbf{x},\textbf{y}) > \epsilon\}$ is open in $X$ under its metric topology. Suppose $\textbf{y} \in A(\textbf{x},\epsilon)$. Then, consider $B_d(\textbf{y},d(\textbf{x},\textbf{y})-\epsilon)$. We show that this open ball lies entirely in $A(\textbf{x},\epsilon)$. Suppose $\textbf{z} \in B_d(\textbf{y},d(\textbf{x},\textbf{y})-\epsilon)$. Then,
$$d(\textbf{x},\textbf{y}) \leq d(\textbf{x},\textbf{z}) + d(\textbf{y},\textbf{z}) < d(\textbf{x},\textbf{z}) + d(\textbf{x},\textbf{y}) - \epsilon$$
implies that $d(\textbf{x},\textbf{z}) > \epsilon$, as desired.

Now we consider the general basis element $(a,b)$ of $\mathbb{R}$. The set $d^{-1}(a,b)$ may be interpreted as the set of all $(x,y) \in X \times X$ such that their distance is greater than $a$ but less than $b$. Then, the set,
$$B = \bigcup _\textbf{x} B_d(\textbf{x}, b)$$
represents the set of all $(x,y) \in X \times X$ whose distance is less than $b$. Clearly, this set is open since it is an arbitrary union of open sets. Similarly,
$$A = \bigcup _\textbf{x} A(\textbf{x}, a)$$
represents the set of all $(x,y) \in X \times X$ whose distance is greater than $a$. This set is open, as we have proven above. Then, $d^{-1}(a,b) = A \cap B$, which is open in $X \times X$.

 

[hedipaldi]

Your proof is correct.Another approach for RxR concerns the metric inducing the topology on RXR.
See attached.
for general metric space use appropriate metric.see answer #3.

 

[hedipaldi]

metric for the producy topology