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Is $f(x,y) = d(x,y)$ a continuous function from a Metric space $X \times X$ to $\mathbb R$? where $(X , d)$ is a metric space.

My Attempt :

I think it is.

If we take any open ball of radius $\epsilon > 0 $ , $(a-\epsilon , a+\epsilon)$ in $\mathbb R$. , We will get an open preimage in $X \times X$.

The preimage will be the union of all $A \times B$ where $A$ and $B $ are open balls of radius $\frac{\epsilon}{2}$ in $X$ whose centers are $a$ units apart.

Have I gone wrong anywhere?

Can anyone please check If I am correct or wrong?

José Carlos Santos
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cmi
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2 Answers2

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Your attempt is interesting, but not correct. Consider the set$$S=\{(x,y)\in X\times X\,|\,d(x,y)=a\}.$$You claimed that$$f^{-1}\bigl((a-\varepsilon,a+\varepsilon)\bigr)=\bigcup_{(x,y)\in S}B_{\frac\varepsilon2}(x)\times B_{\frac\varepsilon2}(y).$$This would indeed prove that $f$ is continuous, but actually you only have$$f^{-1}\bigl((a-\varepsilon,a+\varepsilon)\bigr)\supset\bigcup_{(x,y)\in S}B_{\frac\varepsilon2}(x)\times B_{\frac\varepsilon2}(y).$$I sugest that you prove that $f^{-1}\bigl((a-\varepsilon,a+\varepsilon)\bigr)$ is a neighborhood of all of its points.

José Carlos Santos
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  • I can not understand why you are saying $$f^{-1}\bigl((a-\varepsilon,a+\varepsilon)\bigr)\supset\bigcup_{(x,y)\in S}B_{\frac\varepsilon2}(x)\times B_{\frac\varepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos – cmi Jan 06 '19 at 15:28
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    @cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $\mathbb Q$ with the usual distance and $a=\sqrt2$. Then for every $\varepsilon>0$, the left hand side will contain plenty of pairs. – hmakholm left over Monica Jan 06 '19 at 15:29
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    @HenningMakholm Thank you for saving my time. You are right, of course. – José Carlos Santos Jan 06 '19 at 15:34
  • Is the function uniformly continuous on $X \times X$?@HenningMakholm – cmi Jan 06 '19 at 16:26
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    @cmi Yes, it is. – José Carlos Santos Jan 06 '19 at 16:34
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You want to start from $(x,y)\in X\times X$ and prove that $d$ is continuous at $(x,y)$.

Let $\varepsilon>0$; take $x'\in B(x,\varepsilon/2)$ and $y'\in B(y,\varepsilon/2)$; then $$ d(x',y')\le d(x',x)+d(x,y')\le d(x',x)+d(x,y)+d(y,y')\le d(x,y)+\varepsilon $$ and therefore $$ d(x',y')-d(x,y)\le\varepsilon $$ Can you finish?

Your idea is good as well, but it should be better justified.

vesszabo
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egreg
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