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It just seems like I can't wrap my head around this. For a set $A$ how can I show that the supremum of that set $\sup(A)$ is also a limit point of said set?
Can I just set $$a=\sup(A)$$ $$b\le a \: \: \forall \: b \in A$$ and then show that there's a point different from $a$ in any $\epsilon$-neighbourhood for any $\epsilon>0$? Or should I use a completely different approach?

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    $\begingroup$ Consider $A = \{a\}$. Then $\sup A = a$ is not a limit point of $A$. $\endgroup$ – Daniel Fischer Nov 3 '14 at 14:01
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    $\begingroup$ For future readers, please note that of the five current answers only Xander has pointed out that according to the conventional definition of limit point, @DanielFischer has provided a counter-example to the claim. However, the claim works if we change it to contact point. $\endgroup$ – user21820 Dec 5 '18 at 14:46
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$a$ is the least upper bound for $A$. What does that mean?

First, it's an upper bound. That means each point $b\in A$ satisfies $b\leq a$.

Second, it's the least number that is an upper bound for $A$. This means that any other number $b$ with $b<a$ is not an upper bound for $A$, so there is at least one point $p_b$ of $A$ which exceeds this $b$--that is, $b<p_b$. This means that for each $b<a$, we can demonstrate a $p_b\in A$ with $b<p_b<a$ (the last inequality is due to the fact that $a$ is an upper bound for $A$).

Now define the sequence $b_n=a-\frac1n$. Since $b_n\rightarrow a$, and $b_n<p_{b_n}<a$, we must have that $p_{b_n}\rightarrow a$.

Thus $a$ is a limit point of $A$ since each $p_{b_n}\in A$.

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  • $\begingroup$ this construction of $p_{b_n}$ requires the axiom of choice, doesn't it? $\endgroup$ – Robson Sep 22 '18 at 18:14
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    $\begingroup$ @Robson no. You can just take the least such since $\mathbb{N}$ is well ordered. $\endgroup$ – DRF Dec 6 '18 at 13:21
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    $\begingroup$ @Robson Yes, this requires the axiom of countable choice. The sets $(a-\frac1n,a)\cap A$ are not well ordered. $\endgroup$ – Mike Earnest Dec 16 '18 at 23:45
  • $\begingroup$ argument is wrong. specifically you cannot say that $p_b < a$ all you can say is that $p_b \leq a$.Because upper bounds need not necessarily be strictly greater than elements in set .could someone please confirm ? $\endgroup$ – viru Jun 6 at 7:20
  • $\begingroup$ and definition of limit point specifically asks for sequence where $a_n \neq x$ forall $n \in \mathbb{N}$ for showing x is limit point of set $\endgroup$ – viru Jun 6 at 7:24
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The other answers to this question are incorrect. It is not generally true that the supremum of a set $A$ in $\mathbb{R}$ is a limit point of that set. For example, as pointed out in the comments by Daniel Fischer, for any $a \in \mathbb{R}$, we have $\sup\{a\} = a$, but $a$ is not a limit point of $\{a\}$. To explain this, first recall the definition:

Definition: Let $A \subseteq \mathbb{R}$. A point $x\in\mathbb{R}$ is called a limit point of $A$ if for any $\varepsilon > 0$ there exists some $y\ne x$ such that $|y-x| < \varepsilon$ and $y\in A$. In other words, every $\varepsilon$-neighborhood of $x$ contains a point of $A$ which is different from $x$.

If we take $A = \{a\}$ then, as noted above $\sup(A) = a$. But no neighborhood of $a$ contains any point of $A$ other than $a$. Hence, by the definition above, $a = \sup A$ is not a limit point of $A$. Indeed, given any set $A$ such that $\sup(A) < \infty$, we can construct a set $A'$ by appending an extra point that is greater than the supremum. For example, for any $\varepsilon > 0$, the set $$A' = A \cup \{ \sup(A) + \varepsilon \} $$ satisfies the properties

  1. $\sup(A') = \sup(A) + \varepsilon$, and
  2. $\sup(A')$ is not a limit point of $A'$.

If, instead we ask "Is the supremum of a subset of $\mathbb{R}$ an adherent point of $A$?", then the answer is "Yes", subject to appropriate caveats.

Definition: Let $A \subseteq \mathbb{R}$. A point $x\in\mathbb{R}$ is called an adherent point of $A$ if for any $\varepsilon > 0$ there exists some $y$ such that $|y-x| < \varepsilon$ and $y\in A$. In other words, every $\varepsilon$-neighborhood of $x$ contains a point of $A$.

Note here that we do not require $y$ to differ from $x$. Hence $a$ is an adherent point of $\{a\}$. Note that every limit point of a set is also an adherent point of that set, and that all the points of $A$ are also adherent points. To show that the supremum of a bounded-from-above set is an adherent point, we can argue as follows:

  • If $\sup(A) \in A$, then $\sup(A)$ is an adherent point of $A$, and we are done.
  • If $\sup(A) \not\in A$, then it follows from the definition of the supremum that for any $\varepsilon > 0$ there is a point $$x_n \in (\sup(A)-\varepsilon, \sup(A)]. $$ In other words, for any $\varepsilon > 0$, we can find a point $x_n \in A$ such that $|x_n - \sup(A)| = \sup(A) - x_n < \varepsilon$. This implies that $\sup(A)$ is an adherent point of $A$.

Note that the second case is (in every way that matters) the argument given by MPW, mookid, and Paul in their answers.

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Hint: let $n>0$. Then there is $x_n\in A$ such as $$ \sup A - \frac 1n < x_n \le \sup A $$

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  • $\begingroup$ It is a nice answer. $\endgroup$ – Paul Nov 3 '14 at 13:52
  • $\begingroup$ No. definition of limit point specifically asks for sequence where $a_n \neq x$ forall $n \in \mathbb{N}$ for showing x is limit point of set $\endgroup$ – viru Jun 6 at 7:29
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You can construct a senquence which converges to $a$.

For $a-1$, then there exists $a_1 \in A$ such that $a-1<a_1<a$.

Then for $a_1<a$, there exists $a_2 \in A$ such that $\max\{a_1, a-\frac12\}<a_2<a$

Then for $a_2<a$, there exists $a_3 \in A$ such that $\max\{a_2, a-\frac13\}<a_2<a$

$\cdots$

Then there is a sequence $\{a_n\}\in A$ which converges $a$.

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The assertion is only true for open sets in R. Any neighborhood of supA except itself intersects A must be nonempty otherwise there is a smaller upper bound in that region. Also supA is not in A.

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    $\begingroup$ This is not correct. $\sup([0,1]) = 1$, but the closed interval $[0,1]$ is, well, closed. Not open. $\endgroup$ – Xander Henderson Dec 5 '18 at 5:27
  • $\begingroup$ The assertion supremum is a limit point is wrong, for sets not open. $\endgroup$ – james0910 Dec 5 '18 at 5:40
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    $\begingroup$ Your example shows supremum can be limit points for sets that are not open. But it is not true for all those sets. @XanderHenderson $\endgroup$ – james0910 Dec 5 '18 at 5:47
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    $\begingroup$ The assertion in the original question is that the supremum of a set is a limit point of that set. You claim that this is "only true for open sets." Your claim is incorrect, as demonstrated by my counterexample. If your intention is to state that the supremum of an open set is a limit point, you would be correct (though this doesn't answer the question). However, there are other sets for which the supremum is a limit point---for example, the set $\{-1/n : n\in\mathbb{N}\}$, which is neither open nor closed, and which has 0 as its supremum and a limit point. $\endgroup$ – Xander Henderson Dec 5 '18 at 14:01
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    $\begingroup$ James, you have asserted that the claim "$\sup(A)$ is a limit point of $A$" is true only if $A$ is open. That is the correct literal reading of the answer which you have provided. This statement is false. The claim happens to be true for any bounded open set, but there are non-open sets $A$ such that $\sup(A)$ is a limit point of $A$. I agree that there likely is not a nice, non-tautological description of all of the sets with the property, but your assertion, as written in your answer, is incorrect. $\endgroup$ – Xander Henderson Dec 5 '18 at 15:35

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