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Let $(E, d)$ be a metric space, $x$ element of $E$. Show that the set \begin{equation} A = \{y \in\ E : d(x, y) \geq 5 \} \end{equation} is closed.

Generally, how would you go about this?

I have an exam soon and I want to learn this.

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    okay, doing that. Thank you. – 2012-12-15

4 Answers 4

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Let $z\in E-A$. Then $d(x,z)<5$. Consider the open ball $B$ with centre $z$ and radius $5-d(x,z)$. For any $b\in B$, we have $d(b,x)\leq d(b,z)+d(z,x)<5-d(x,z)+d(z,x)=5$. This shows that $E-A$ is open so that $A$ is closed.

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    how does "This shows that $Eāˆ’A$ is open" mean that "$A$ is closed" ? Is it because *A set is closed iff its complement is open.* ? – 2018-04-14
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Use that $d: E \times E \to \mathbb R_{\ge 0}$ is continuous in each argument and that $A = d^{-1}(x, [5, \infty)) = d_x^{-1}([5, \infty))$ where $[5,\infty)$ is a closed set and $d_x (y) = d(x,y)$ is a continuous function.

More generally, use that the inverse of a continuous function maps open sets to open sets and closed sets to closed sets.

You can use this to see why $f^{-1}(\{a\}) = A$ is closed.

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    @user108903 Yes. That's right. Better? – 2012-12-15
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The function $f:E\to R$ defined by $f(y)=d(y,x)$ is continuous. Since $[5,\infty)$ is closed in $R$, so $A=f^{-1}([5,\infty))$ closed.

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Choose $p\notin A$. Let $r=\min${$d(x,p),5-d(x,p)$}$\implies B(p,r)\cap A=\emptyset\implies p\notin cl(A)$