Can someone please verify my proof or offer suggestions for improvement?

I am aware that there is a similar question elsewhere, but I want help with my proof in particular.

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

Let $U \subseteq \mathbb{R}$ be open. Assume $\operatorname{d}^{-1}(U) \neq \varnothing$.

Let $(x, y) \in \operatorname{d}^{-1}(U)$ and set $\epsilon_1 = \frac{1}{2}(b-\operatorname{d}(x, y))$

Clearly, if $(\alpha, \beta) \in B_d(x, \epsilon_1) \times B_d(y, \epsilon_1)$, then \begin{eqnarray} \operatorname{d}(\alpha, \beta) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(x, y) + \operatorname{d}(y, \beta)\\ &<& \operatorname{d}(x, y) + 2\left(\frac{b-\operatorname{d}(x, y)}{2}\right) \\ \end{eqnarray}

That is,

$$\operatorname{d}(\alpha, \beta) < b$$

Similarly, set $\epsilon_2 = \frac{1}{2}(\operatorname{d}(x, y) - a)$.

Let $(\alpha, \beta) \in B_d(x, \epsilon_2) \times B_d(y, \epsilon_2)$. Assume, for the sake of contradiction, that $\operatorname{d}(\alpha, \beta) \leq a$.


\begin{eqnarray} \operatorname{d}(x, y) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(\alpha, \beta) + \operatorname{d}(\beta, y) \\ \operatorname{d}(x, y) &<& \operatorname{d}(x, y)-a + \operatorname{d}(\alpha, \beta) \\ &<& \operatorname{d}(x, y) \end{eqnarray}

Therefore, it must be the case that $$\operatorname{d}(\alpha, \beta) > a$$

Set $\epsilon = \operatorname{min}\{\epsilon_1, \epsilon_2\}$.

Then, if $(\alpha, \beta) \in B_d(x, \epsilon) \times B_d(y, \epsilon)$,

$$a < \operatorname{d}(\alpha, \beta) < b$$

Since $\operatorname{d}^{-1}(U)$ can be written as the union of open sets in the metric topology, it follows that $\operatorname{d}^{-1}(U)$ is open. Therefore, $d$ is continuous.

  • $\begingroup$ You never say what $a$ and $b$ are, and you show that $d^{-1}(U)$ is open only for open intervals [the interval $(a,b)$, as transpires], not for general open sets. If you state that you assume that $U = (a,b)$ at the beginning of the argument, and then (at some point) say that the general case follows, since every open subset of $\mathbb{R}$ is a union of such intervals, and say that the conclusion is clear if $d^{-1}((a,b)) = \varnothing$, then your proof will be correct. $\endgroup$ Aug 10, 2014 at 16:52
  • $\begingroup$ Oh. If I clarify what $a$ and $b$ are, and state that all intervals of the form $(a,b)$ are basis elements for the standard topology on $\mathbb{R}$, will the proof be correct? $\endgroup$
    – user154185
    Aug 10, 2014 at 17:00
  • $\begingroup$ Well, since you say "Assume $d^{-1}(U) \neq \varnothing$", you also need to say what the matter is in case that is empty. $\endgroup$ Aug 10, 2014 at 17:03

3 Answers 3


In fact you proved that $d^{-1}((a,b))$ is open for any interval $(a,b)\subset \mathbb{R}$.

Since any open set $U$ can be written as union of some family of intervals $\{(a_i,b_i):i\in I\}$, then $$ d^{-1}(U)=d^{-1}(\cup_{i\in I}(a_i,b_i))=\cup_{i\in I}d^{-1}((a_i,b_i)) $$ Thus $d^{-1}(U)$ is open as union of open sets.


It is easy by using the following inequality : $|d(x,y)-d(x',y')|\leq d(x,x')+d(y,y')$.


What you are looking to use is a Product Metric


Using $p=1$ is just fine. It gives you a metric for your product space based on your previous metrics, like so: If $d'$ is your new metric for $XxX$ with and using metric $d$ on $X$,

$$ d'((x_1,y_1),(x_2,y_2)) = d(x_1,y_1)+d(x_2,y_2)$$

Then, all you have to do is prove that metrics are continuous, which is pretty easy to do.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.