Given a metric space $(X,d)$, how to prove that the function $d \colon X \times X \to \mathbf{R}$ is continuous?

If we take any two arbitrary real numbers $a$ and $b$ such that $a < b$, then we need to show that the set $d^{-1} (a,b)$ given by

$$ d^{-1} (a,b) := \{ (x, y) \in X \times X | a < d(x,y) < b \} $$

is open in the product topology on $X \times X$.

A basis for this product topology is the collection of all cartesian products of open balls in $(X,d)$.


2 Answers 2


For $a, b ∈ ℝ$, let $(x,y) ∈ d^{-1} (a..b)$, i.e. $a < d(x,y) < b$. Now choose $ε$ such that $U_{2ε} (d(x,y)) ⊂ (a..b)$ and look at $U_ε (x) × U_ε (y)$.

For any tuple of points $(x',y') ∈ U_ε (x) × U_ε (y)$ you have $$d(x',y') ≤ d(x',x) + d(x,y) + d(y,y') < d(x,y) + 2ε$$ as well as $$d(x,y) ≤ d(x,x') + d(x',y') + d(y',y) < d(x',y') + 2ε$$ This means $a < d(x,y) - 2ε < d(x',y') < d(x,y) + 2ε < b$.

Therefore $U_{ε} (x) × U_{ε} (y) ⊂ d^{-1} (a..b)$.

  • $\begingroup$ Sorry, I didn't recognize the notation, but by $(a..b)$ do you mean the set of all real numbers from $a$ to $b$ exclusive (i.e. the open interval)? $\endgroup$ Jul 16, 2016 at 21:07
  • 4
    $\begingroup$ Yeah, it’s a notation introduced by Knuth, I believe. It’s quite handy. It avoids the confusion of regarding $(a,b)$ both as a tuple and as an open interval. See the good notations thread on mathoverflow. $\endgroup$
    – k.stm
    Jul 17, 2016 at 17:53
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    $\begingroup$ So in the metric space $X$, we can take any point $x, y$ such that $d(x,y)=k$ with any given $k \in R$ ? $\endgroup$
    – RopuToran
    Jan 2, 2021 at 17:21
  • $\begingroup$ @RopuToran No, why would you think that? In ℚ, you find no two points with distance exactly $\sqrt 2$. $\endgroup$
    – k.stm
    Jan 3, 2021 at 13:28

Let $(x,y)\in d^{-1}(a,b)$. Define $\epsilon=\frac{1}{100}\min(d(x,y)-a,b-d(x,y))$. Then for any point $(\xi,\gamma)\in B(x,\epsilon)\times B(y,\epsilon)$, we have $${\color{blue} d(\xi,\gamma) }\leq d(\xi,x)+d(x,y)+d(y,\gamma)<2\epsilon+d(x,y)<{\color{blue}b}$$ and $${\color{blue} a}<d(x,y)-2\epsilon <d(\xi,\gamma)+\underbrace{ d(x,\xi)-\epsilon}_{<0} +\underbrace{d(y,\gamma)-\epsilon}_{<0}<{\color{blue} d(\xi,\gamma)}.$$ That is to say $$(x,y)\in B(x,\epsilon)\times B(y,\epsilon)\subseteq d^{-1}(a,b).$$

Hence any point $(x,y)\in d^{-1}(a,b)\subseteq X\times X$ is surrounded by some open set contained in $d^{-1}(a,b)$.

(After typing this answer, I realized this is EXACTLY the same as the answer by @k.stm; so thanks also to his contribution. I guess it will be hard to write a different one.)


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