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Let $A\subset\mathbb R$ be a set.We know that $\inf(A)$ is a member of the closure of $A$. We know that $\inf(A)$ is the greatest lower bound for the set $A$.

Q1) Is it always true that $\inf(A)=\liminf{a_n}$ for some sequence $a_n$ formed by the points in $A$?

Q2) $\text {diam}(A)=\limsup|p_n-q_n|$ for two sequences of $p_n$ and $q_n$ right? So if $p_n,q_n$ does not converge, Since they are bounded and in $\mathbb R$ their subsequences converge. How do we know that their subsequences converge to $\sup$ but not $\inf$ while trying to find $\text {diam}(A)$?

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    $\begingroup$ Q1 is obviously true. You said that $\inf(A)$ is a member of the closure of $A$, thus it is a limit point of $A$. Question Q2 I cannot understand $\endgroup$ – Matias Heikkilä Oct 9 '16 at 12:59
  • $\begingroup$ @MatiasHeikkilä: it does not have to be a limit point of $A$. If $A = \{\inf(A)\}$, then it is an isolated point of $A$. $\endgroup$ – user87690 Oct 9 '16 at 13:17
  • $\begingroup$ Yes, but then you can take $x_n = \inf(A)$. $\endgroup$ – Matias Heikkilä Oct 9 '16 at 13:19
  • $\begingroup$ As Matias pointed out, Q1 is obviously true. About Q2, I don't understand why you work with limit superior, instead of limit: By definition, $\text{diam} A=\sup\{|p-q|: p, q\in A\}$, so there exist $(p_n)$, $(q_n)$ in $A$ such that $|p_n-q_n|\rightarrow \text{diam} A$. Any subsequence of them also has the same property. This follows immediately from the definition of a subsequence. $\endgroup$ – tree detective Oct 9 '16 at 13:20
  • $\begingroup$ @MatiasHeikkilä: Yes, of course. “a member of closure $\implies$ a limit of some sequence” is true, but “a member of closure $\implies$ a limit point of the set” is not. $\endgroup$ – user87690 Oct 9 '16 at 13:22
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Let $A\subset\mathbb R$ be a set. We know that $\inf(A)$ is a member of the closure of $A$.

This is incorrect. Consider for instance $A=\mathbb{R}$. If one assumes that $A$ is bounded, then it would be true.

Q1) Is it always true that $\inf(A)=\liminf{a_n}$ for some sequence $a_n$ formed by the points in $A$?

Yes, this is true by the definition of infimum. Actually, you might have $$ \inf(A)=\lim a_n $$ for some sequence $a_n$ in $A$.

Q2) $\text {diam}(A)=\limsup|p_n-q_n|$ for some sequences $p_n$ and $q_n$, right?

Yes. You can find the two sequences by using the definition of "$\sup$" and actually you could have $$\text {diam}(A)=\lim|p_n-q_n|.$$

So if $p_n,q_n$ does not converge, since they are bounded and in $\mathbb R$ their subsequences converge.

This does not make sense and it is logically wrong.


There seems to be some confusion about the relation between $\inf$ and $\liminf$, and the relation between $\sup$ and $\limsup$. Here is a simpler helpful exercise:

  • Suppose $A$ is a subset of $\mathbb{R}$ and let $a=\inf A$, $b=\sup A$ where $a,b$ can be extended real numbers. Show that there exists sequences $a_n$ and $b_n$ in $A$ such that $$ \lim_na_n=a,\quad \lim_nb_n=b. $$
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  • $\begingroup$ does $inf(R)$ exist? Do you assume the set to be compact when saying$diam(A)=\lim|p_n-q_n|$ $\endgroup$ – user346936 Oct 9 '16 at 16:12
  • $\begingroup$ (1) Well, I'm assuming you use the extended real numbers and thus $\inf(\mathbb{R})=-\infty$. (2)No, I don't even assume that $A$ is bounded. $\endgroup$ – Jack Oct 9 '16 at 16:15

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