Let $(X, d)$ be a metric space. Is the function $x\mapsto d(x, z)$ continuous? Is it uniformly continuous?
Is the distance function in a metric space (uniformly) continuous?
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1@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 – 2013-06-25
2 Answers
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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1@GAJO I think $d(x,E)$ is continuous in $x$ for any nonempty subset $E$. – 2014-10-14
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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4Isn't it also the coarsest one for which $d$ itself is continuous? – 2010-10-27