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Let $(X,\rho)$ be a metric space. Show that $\rho(x,E) > 0 $ where $E$ is compact and $x \notin E $

I dont think we can assume that $E$ is closed since this is a arbitrary metric space.

My attempt

1) consider $f(x) = \inf\{\rho(x,y): y \in E\}$. Then by extreme value theorem $f(x) \geq 0$. How to show that $f(x) \neq 0$.

Any other method is also welcome

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    Let $K(x,n) = \{y \in X| d(x,y) > 1/n\}$ then $\cup_{n\in \mathbb N}K(x,n)$ is an open cover of $E \subset X-\{x\}$. So it has a finite subcover. So $E \subset \cup_{n \in L}K(x,n)$ where L is a finite set of natural numbers. let k = max L. Then all $d(x,y) > 1/k$ for all y in E so d(x,E) > 1/k > 0. – 2017-01-06

3 Answers 3

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The function defined on $E$ by $f(y)=d(x,y)$ is continue, since $E$ is compact, there exists $y_0$ such that $f(y_0)=min_{y\in E}f(y)=d(x,E)=d(x,y_0)>0$ since $x\neq y_0$.

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    How minimum is $>$ than 0 that is my question. it can be = $0$ !!! – 2017-01-06
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    because $min=f(y_0)=d(x,y_0)$ if $x\neq y_0$, $d(x,y_0)\neq 0$ property of a metric. $x\neq y_0$ since $x$ is not in $E$ and $y_0\in E$. – 2017-01-06
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    :(....I was so close....Thanks man – 2017-01-06
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    But do you think that we can assume that E is closed? – 2017-01-06
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    $E$ is compact in a metric space, thus closed, but closed is not enough to have this property. – 2017-01-06
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    this result is true. – 2017-01-06
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51327/discussion-between-silent-learner-and-tsemo-aristide). – 2017-01-06
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Towards a contradiction suppose that $d(x,E)=0$. Then there exists a sequence of points $(x_n)$ in $E$ such that $x_n\to x$. But $E$ is closed.

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    How can we assume that E is closed?... E might not be closed – 2017-01-06
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    @silentlearner Nope, compact sets have to be closed. If there's a sequence converging to a point outside the set, then that's also an example of a sequence with no accumulation point inside the set. – 2017-01-06
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    Every compact subset of a metric space is closed. For arbitrary topological spaces, the result may not be true. I – 2017-01-06
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    ^Metric spaces are Hausdorff so compact subsets are closed. – 2017-01-06
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    Note indeed that compactness is not needed, just closedness which follows from it, Compactness does imply that there is some $y \in E$ such that $\rho(x,E). =\rho(x,y)$ – 2017-01-06
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Since $E$ is a compact subseet of a metric space, it is closed. Therefore $E^c$ is open, so if $x\not\in E$ we can choose $r> 0$ so that $B_r(x)\subseteq E^c$. Hence $d(x,E)\ge r$.