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Let $X$ be a metric space , $d$ is the metric , show that $d$ is a continuous function from $X\times X$ to $R$.

I think the definition is all we need , but I just don't know where to start , can anyone help me.

Nitin Uniyal
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Idele
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2 Answers2

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Define $f:X\times X\to \mathbb R$ by $f(x,y)=d(x,y)$.

Let $(x_n,y_n)$ be a sequence in $X\times X$ such that $(x_n,y_n)\to (x,y)$

Then $(x_n)\to x,(y_n)\to y\implies d(x_n,x)\to 0 ,d(y_n,y)\to 0 $ as $n\to \infty\implies d(x_n,y_n)\leq d(x_n,x)+d(x,y)+d(y_n,y)\to d(x,y)$

$\implies d(x_n,y_n)\to d(x,y) $ as $n\to \infty \implies f(x_n,y_n)\to f(x,y)$

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    $d(x_n,y_n)\leq$ something, that converges to $d(x,y)$ how it implies that it's limit is that only @Learnmore – Devendra Singh Rana Jun 19 '18 at 10:53
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    @DevendraSinghRana He missed a detail. Use triangle inequality to $d(x,y)$ to obtain that the limit of $d(x_n,y_n)$ is greater or equal $d(x,y)$. Then you obtain the asymptotic equality. – Celine Harumi Jul 11 '19 at 04:57
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    @Devendra Singh rana you can imagine it like a quadrilateral whose breath converges to 0 – Lord Shadow Aug 24 '20 at 15:04
  • There's also the issue here that for 1st countable spaces ( incl Metric Spaces), Sequential Continuity implies continuity. not true for all spaces. – MSIS Sep 11 '22 at 21:02
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Let $\varepsilon > 0$, and let $(x_1,x_2) \in X \times X$. Then if we let $\delta = \frac\varepsilon2$ we get that $U = B_\delta(x_1) \times B_\delta(x_2)$ is a neighborhood of $(x_1,x_2)$ in $X \times X$ such that $d(U) \subseteq (d(x_1,x_2) - \varepsilon, d(x_1,x_2) + \varepsilon)$ by an application of the triangle inequality. Therefore $d:X \times X \to \mathbb R$ is a continuous function.

silvascientist
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    Could you explain to me that 'therefore'? I don't see the implication... Thank you! – John Mars Dec 24 '20 at 18:13
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    @JohnMars A real-valued function $d$ is continuous at an element $x$ of its domain if, for every $\varepsilon > 0$, there is a neighborhood $U$ of $x$ such that $d$ maps $U$ into $(d(x) - \varepsilon, d(x) + \varepsilon)$, which is exactly what was shown, where $x = (x_1,x_2)$ – silvascientist Dec 25 '20 at 00:43
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    Oh ok, there is a neighborhood... here was my doubt. – John Mars Dec 25 '20 at 01:15