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let $A$ be a nonempty subset of a metric space $M$. If $\epsilon\ge0,$ show that the set of $x\in M$ s.t $\rho(x,A)\ge\epsilon$ is closed.

let $E=\{x\in M | \rho(x,A)\ge\epsilon\}$

I am thinking of three different ways to prove this:

1- by showing $E=\bar{E}$. which is ta king a point in $E=\bar{E}$ and show that point also belongs in $E$. ( the other direction is clearly true )

2- by showing $E^c$ is open.

3-by using the concept of functions.

I believe I can show the proof for methods 1 and 2. However, I have no idea how to use the concept of a function to prove the statement above. I would appreciate such proof. Thanks.

Clement Yung
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BesMath
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  • The theorem stated and proved here may be of interest https://math.stackexchange.com/questions/437829/the-preimage-of-continuous-function-on-a-closed-set-is-closed – Mark Feb 09 '20 at 15:58

2 Answers2

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I would choose approach 2- here, because the topology of a metric space is generated by the set of all open balls $\{B(x,r)\ |\ x\in M,r\in (0,\infty)\}$ in $M$ as their subbasis, where $$B(x,r):=\{y\in M\ |\ \rho(x,y)< r\},$$ i.e. all (finite or infinite) unions of these balls are open by definition of the topology.

Now verify that $$E^c=\bigcup_{x\in A}B(x,\varepsilon)$$ and you have a representation of $E^c$ as such a union of open balls, meaning $E^c$ is open.

Formyer
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The function $ f: X \to \Bbb R$ defined by $f(x)=d(x,A)$ is continuous (e.g. because we can show that $|f(x)-f(y)|\le d(x,y)$ follows from the triangle inequality).

Your set is then $f^{-1}[[\varepsilon, +\infty)]$, an inverse image of a closed set under a continuous function and thus closed.

Henno Brandsma
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