# Show that the set is closed

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

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

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$A=E\setminus B_d(x,5)$, where $B_d(x,5)=\{y\in E:d(x,y)<5\}$. $B_d(x,5)$ is an open ball, so ... ? –  Brian M. Scott Dec 15 '12 at 11:33
okay, doing that. Thank you. –  MathMathMath Dec 15 '12 at 11:57

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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+1. Although I already have "Sportsmanship" : ) –  Matt N. Dec 15 '12 at 11:41
Great answer. Makes complete sense to a beginner like me :) Can you also try answering my new added question? –  MathMathMath Dec 15 '12 at 11:53

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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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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$d^{-1}(x,[5,\infty))$ looks odd to me. I guess you mean $d_x^{-1}([5,\infty))$ where $d_x(y)=d(x,y)$. –  user108903 Dec 15 '12 at 11:43
@user108903 Yes. That's right. Better? –  Matt N. Dec 15 '12 at 11:45
Choose $p\notin A$. Let $r=\min${$d(x,p),5-d(x,p)$}$\implies B(p,r)\cap A=\emptyset\implies p\notin cl(A)$