$A\subseteq\mathbb{R}$ is closed iff $\sup(A\cap[a,b])\in A$ and $\inf(A\cap[a,b])\in A$.

Question

Suppose that $A\subseteq\mathbb{R}$. Show that the following are equivalent:

(a) $A$ is closed.

(b) If $[a,b]$ is a closed interval for which $A\cap[a,b]$ is non-empty, then $\sup(A\cap[a,b])\in A$ and $\inf(A\cap[a,b])\in A$.

MY ATTEMPT

Let us prove the implication $(a)\Rightarrow(b)$ first.

Since $A\cap[a,b]$ is non-empty and bounded above by $b$, it admits a supremum.

Moroever, we do also have that $A\cap[a,b]\subseteq A$. Hence we conclude \begin{align*} \sup(A\cap[a,b])\in\overline{A\cap[a,b]}\subseteq\overline{A} = A \Rightarrow \sup(A\cap[a,b])\in A \end{align*}

This is because the supremum of a set is a closure point of it. Similar reasoning applies to the infimum.

However I do not know how to approach the converse implication.

Could someone tell me what I am missing?

Any contribution is appreciated.

Answer 1

Hint: Suppose $A$ is not closed. Then, there exists some converging (and WLOG monotonic) sequence $x_1,x_2,\dots\in A\subset \mathbb{R}$ such that $\lim_{n\to \infty}x_n=x\not\in A$. Take $[a,b]=[x-1,x]$ if the sequence is increasing, or $[a,b]=[x,x+1]$ if it is deceasing, and observe...

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