I'd appreciate some input in how I can (appropriately) simplify the reasoning in the following simple proof:

Let $E \subset \mathbb{R}$ be non-empty and bounded above. Let $y =$ sup $E$.

Prove $y$ is in the closure of $E$.

Proof: Let $\overline{E} = E \cup E'$ denote the closure of $E$, where $E'$ is the set of all limit points of $E$.

If $y \in E$, then clearly $y \in \overline{E}$.

Suppose $y \notin E$. For all $r > 0$, there must be an $x \in E$ such that $y - r < x < y$ since if it were not the case, $y - r$ would be a lower bound, contradicting $y$ being the least upper bound. Therefore we have a neighborhood centered at $y$ with an arbitrary radius $r > 0$ that contains $x \in E$. Therefore $y$ is a limit point of $E$, so $y \in E'$ and thus $y \in \overline{E}$. $\blacksquare$

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    $\begingroup$ Generally I would prefer constructive proofs where possible. If $\sup E \in E$ there is nothing to do, so suppose $\sup E \notin E$. By definition of $\sup E$, for all $n$ there is some $x_n \in E$ such that $x_n > \sup E - {1 \over n}$. Then $x_n \to \sup E$, and hence $\sup E \in E'$. $\endgroup$
    – copper.hat
    Jul 27, 2018 at 21:45
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    $\begingroup$ I provided a constructive proof for the inf on math.stackexchange.com/questions/2852981/…. E.g. I proved there's a sequence that converges to it, so it's a limit point, so it's in the closure. $\endgroup$
    – user518441
    Jul 27, 2018 at 21:48
  • $\begingroup$ I'm actually fine with this kind of proof. All I would really change is the wording in one part of it: say, "Therefore, for $r>0$, $B_r (y)\neq\emptyset$, so $y\in E'$." The way you've worded it as it stands is a bit redundant. $\endgroup$ Aug 8, 2020 at 1:54

1 Answer 1


The aim is to construct a sequence $(a_n)$ in $A$ such that it converges to the supremum. Then the supremum is a limit point pf $A$ and by definition of closure, it will belong to $\overline{A}$.


Let $\operatorname{sup}(A)=a$. Since $a$ is the supremum of $A$, for $\epsilon>0$, there is an element $x\in A$ such that $a-\epsilon<x$. For the particular case that $\epsilon=1/n$ ($n$ integer), we can define a sequence $(a_n)$ in $A$ such that

$$ a - 1/n < a_n < a. $$

By the Squeeze theorem, $(a_n)\rightarrow a$. Thus, $a$ is a limit point of $A$ and $a\in\overline{A}$.

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    $\begingroup$ The only contention I might have with this proof is that in some introductory analysis courses (which I would assume of the author based on the question being asked), you're asked to prove these kinds of things without things like the Squeeze Theorem. Fortunately, the fact that you've constructed a sequence like this doesn't require the Squeeze Theorem; the construction itself shows that for any radius $\varepsilon>0$, $B_\varepsilon (a)\neq\emptyset$, satisfying the definition of a cluster/limit point. $\endgroup$ Aug 8, 2020 at 1:47
  • $\begingroup$ If I remember correctly, the question by the OP is based in the book by Abbot, Understading Analysis. In this case, if my memory doesn't fail me, these discussions on topology (Chapter 3) come afterwards limit and sequences (Chapter 2). That's why I maliciously used the trick of the Squeeze Theorem. $\endgroup$ Aug 9, 2020 at 17:50

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