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Let $(X, d)$ be a metric space. Is the function $x\mapsto d(x, z)$ continuous? Is it uniformly continuous?

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    $\begingroup$ If you are asking whether the distance function $d:X\times X\to\mathbb R$ is continuous when $(X,d)$ is a metric space, then aswer is yes. You should try to prove it yourself; first do it for $\mathbb R$ with its usual metric, and then generalize. $\endgroup$ – Mariano Suárez-Álvarez Oct 27 '10 at 16:32
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    $\begingroup$ @MarianoSuárez-Alvarez I was also trying the same question. But how to prove it using $\epsilon - \delta$ notation.Let $(x,y) \in X$ and $(x',y') \in X$ then whenever $||(x,y)-(x',y')|| \lt \delta$ then we have $|d(x,y)-d(x',y')| \lt \epsilon$.. right?? How to go ahead with this proof $\endgroup$ – Mathy Jun 25 '13 at 11:33
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As Qiaochu points out $d(x,y)$ is continuous for fixed $x$. You may like to see this as well, as this is a familiar result in Topology:

If $A$ is a non empty subset of a metric space $(X,d)$ then the function $f$ on $X$ given by $$f(x)=d(x,A):= \inf_{y\in A} d(x, y)$$ is continuous. Indeed, $$| f(x) - f(y) | = | d(x,A) - d(y,A) | \leq d(x,y),$$ and thus $f$ is uniformly continuous (use $\delta = \epsilon$ in any point).

To show this, let $x$ and $y$ be points in $X$, and $p$ any point in $A$.

Then $$d(x,p) \leq d(x,y) + d(y,p)\ \ \ \ \text{ (triangle inequality)}$$ and so $$d(x,A) \leq d(x,y) + d(y,p)$$ as $d(x,A)$ is the infimum. But then $d(y,p) \geq d(x,A) - d(x,y)$ (for all $p$, obtained by subtracting from the previous inequality), so that $d(y,A) \geq d(x,A) - d(x,y)$ (as $d(y,A)$ is the infimum). So : $d(x,A) - d(y,A) \leq d(x,y)$.

Now reverse the roles of $x$ and $y$ to get $d(y,A) - d(x,A) \leq d(x,y)$.

This is taken from http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2004;task=show_msg;msg=1323.0001

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    $\begingroup$ where did you use the fact that A is closed? $\endgroup$ – GAJO Jan 28 '14 at 19:11
  • $\begingroup$ Why are we assuming that the metric on $\mathbb{R}$ is the standard one? $\endgroup$ – The Substitute Aug 31 '14 at 0:19
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    $\begingroup$ @GAJO I think $d(x,E)$ is continuous in $x$ for any nonempty subset $E$. $\endgroup$ – Fang Jing Oct 14 '14 at 20:29
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Yes. The standard definition of the topology induced by a metric ensures this; in fact it's not hard to see that it's the coarsest topology such that $d(x, y)$ is continuous for fixed $x$.

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    $\begingroup$ Isn't it also the coarsest one for which $d$ itself is continuous? $\endgroup$ – Mariano Suárez-Álvarez Oct 27 '10 at 16:36

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