Consider A⊂RA\subset \mathbb{R}, where AA is bounded. Since AA is bounded, sup(A)\sup(A) exists. To show that sup(A)∈Closure(A)\sup(A)\in Closure(A), we need to show that sup(A)\sup(A) is a limit point of AA. Let sup(A)=s∈R\sup(A)=s\in\mathbb{R}. If AA is a finite set then it is closed, so the closure of AA is AA, and sup(A)∈Closure(A)\sup(A)\in Closure(A). If AA is not finite, consider the sequence {s−1/n}∩A\{s-1/n\}\cap A. This sequence converges to ss and is entirely contained in AA.

But I have some doubts about my proof approach because if we consider A:=(0,1/2)∪{5}A:=(0,1/2)\cup\{5\}, the above sequence will not intersect AA. In this case, however, one can argue that {5}\{5\} is a singleton, which is closed, and so {5}\{5\} must be in the closure of AA. In all other cases, it seems, the above sequence will be contained in AA and will converge to ss. Is this true?

Another approach I could use is this. Suppose s∉Closure(A)s\not\in Closure(A). Then s∈R∖Closure(A)s\in \mathbb{R}\backslash Closure(A). Since Closure(A)Closure(A) is closed, R∖Closure(A)\mathbb{R}\backslash Closure(A) is open, thus ∃\exists an r>0r>0 such that the open ball Br(s)∩̸Closure(A)B_r(s)\not\cap Closure(A). Which is impossible if ss is not a singleton. If ss is a singleton, then s\in Closure(A)s\in Closure(A) by definition. Therefore, s\in Closure(A)s\in Closure(A) in all cases.

Please let me know about the two of my approaches.

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The intersection \{s – 1/n \} \cap A\{s – 1/n \} \cap A can just be empty.

– xyzzyz

Oct 20 at 23:41

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3 Answers

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To show that a real number xx lies in the closure \bar{A}\bar{A} of AA, you have to show that either x\in Ax\in A or that xx is a limit point of AA. While some points in AA are limit points of AA, your example A=(0,\frac{1}{2})\cup\{5\}A=(0,\frac{1}{2})\cup\{5\} demonstrates that not every element of \bar{A}\bar{A} is a limit point of AA, as it could be an isolated point in AA.

So to show that \alpha=\sup(A)\in\bar{A}\alpha=\sup(A)\in\bar{A}, we have to show that either \alpha\in A\alpha\in A or that \alpha\alpha is a limit point of AA. If \alpha\in A\alpha\in A then we’re done, so suppose that \alpha\not\in A\alpha\not\in A.

Since \alpha-1\alpha-1 is not a upper bound for AA, it follows that we can choose some s_1\in As_1\in A such that \alpha\geq s_1>\alpha-1\alpha\geq s_1>\alpha-1. Now s_1\neq \alphas_1\neq \alpha because \alpha\not\in A\alpha\not\in A, so since neither \alpha-\frac{1}{2}\alpha-\frac{1}{2} nor s_1s_1 is an upper bound for AA, we can choose s_2\in As_2\in A such that \alpha \geq s_2>\max\{s_1,\alpha-\frac{1}{2}\}\alpha \geq s_2>\max\{s_1,\alpha-\frac{1}{2}\}.

Proceeding inductively, we can construct a sequence \{s_n\}\{s_n\} such that \alpha\geq s_n>\max\{s_{n-1},\alpha-\frac{1}{n}\}\alpha\geq s_n>\max\{s_{n-1},\alpha-\frac{1}{n}\} for all n\geq 2n\geq 2. This means that \{s_n\}\{s_n\} is a sequence of distinct elements of AA which converges to \alpha\alpha, hence \alpha\alpha is a limit point of AA.

The definition of the closure of AA is that it contains all limit points of AA. But aren’t all points of AA also its limit points (not necessarily all its limit points)?

– sequence

Oct 21 at 0:20

No, not if you require a limit point to be the limit of a sequence of distinct elements of AA (which is the definition I’m familiar with). For instance, 55 is not a limit point of (1,\frac{1}{2})\cup\{5\}(1,\frac{1}{2})\cup\{5\}.

– carmichael561

Oct 21 at 0:22

Isn’t \{5\}\{5\} a limit point of the sequence \{5,5,5,…,5\}\{5,5,5,…,5\}?

– sequence

Oct 21 at 0:28

That’s not a sequence of distinct elements.

– carmichael561

Oct 21 at 0:29

I see. Looks like the definition I was given is different.

– sequence

Oct 21 at 0:32

Because A is a bounded subset of \mathbb{R}\mathbb{R}, it has a supremum. Then see this answer. The first part of the proof, which is not shown, assumed the case where y \in\in E.

See this answer: Supremum of closed sets

By definition of sup(A), every open ball(“segment in R”) around sup(A) contains at least one element from A, thus making it by definition belonging to Closure(A).