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Consider the sequence $(a_n)_n$ that is defined by $a_n:=1/n$ and the set $A:=\{a_n \ | \ n \in \mathbb{N}\} \cup \{0\}$. I want to show that $\mathbb{R} \setminus A$ is open in $\mathbb{R}$. I had two ideas to show this.

$(1)$ It should be the case that $\mathbb{R} \setminus A = (-\infty,0) \cup (1, \infty) \cup \bigcup_{n \geq 2} \left(\frac{1}{n+1}, \frac{1}{n}\right)$, so it is in particular a union of open sets and thus open.

$(2)$ Let $x \in \mathbb{R} \setminus A$. Then define $\delta:= \min\{|x-a| \ : \ a \in A\}$ and $B:=\left(x-\frac{\delta}{2},x+\frac{\delta}{2}\right)$. Then $B$ is open with $x \in B$ and $B \subseteq \mathbb{R} \setminus A$. Im not quite sure if the minimum exists here, which is why I am not sure if this would work. If one would not have $0$ in A for example and choose $x=0$, then one would have $\delta=0$ which would also be a problem.

Are these attempts correct?

user103178981
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1 Answers1

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Attempt $(1)$ is very direct, and absolutely correct.

Attempt $(2)$ is incorrect for the reason that you pointed out - in any metric space, you need to already know that $A$ is closed to assume that: $$\delta:=\min\{d(x,a):a\in A\}$$Even exists. And even for $x\in\Bbb R\setminus A$, it could happen that this 'minimum' (really, if $A$ is not closed, in general this would just be an infimum) is zero.

I don't know if you've covered enough content to handle this, but here are two general statements you might like to prove:

  1. Let $(X,\rho)$ be any metric space and $(x_n)_{n\in\Bbb N}$ any convergent sequence in $X$. If $\alpha:=\lim_{n\to\infty}x_n$, then show that: $$\alpha\cup\{x_n:n\in\Bbb N\}$$Is a compact subset of $X$, and in particular closed.

  2. Now let $(x_n)_{n\in\Bbb N}$ be any sequence whatsoever. If: $$A:=\{x\in X:x\text{ is a limit of some subsequence of }(x_n)_{n\in\Bbb N}\}$$Then show that $A\cup\{x_n:n\in\Bbb N\}$ is closed.

These generalise your situation.

FShrike
  • 40,125
  • Remark for the proposer: For (2) it suffices that any $x\in \Bbb R$ \ $A$ has a nbhd $U$ such that $U\cap A$ is finite, so $x$ has a nbhd $V$ with $V\subset U$ and $V\cap A=\emptyset$, so $x\not\in\overline A$.... Let $\epsilon=(1/2)d(x,\alpha).$ There are only finitely many $n$ for which $d(x_n,\alpha)\ge\epsilon,$ so let $U=B_d(x,\epsilon)={y: d(y,x)<\epsilon}.$ – DanielWainfleet Nov 15 '22 at 19:24