# Show that infimum of set is in closure of set.

Given some set $S$, and that $u = \inf(S)$, show that $u \in \bar{S}$

I figured that you can say that

\begin{align} & \inf(S) = u \\ & \Rightarrow \forall x \in S, u \lt x \\ & \Rightarrow \forall \varepsilon \gt 0, u - \varepsilon \notin S \ \land u + \varepsilon \in S \end{align}

and that implies that $\inf(S)$ is a boundary point of $S$, and thus will be in $\bar{S}$. Does that make sense, or is there a better way to show it?

• $2$nd step does not imply the $3$rd one. Instead, think about how you can create a sequence in S with a limit point $u$. Apr 24, 2017 at 12:42
• @Pawel hmm well by definition $\inf(S)$ is not in the set, and $\forall \varepsilon \gt 0, u + \varepsilon$ will be in the set again by definition of the $\inf(S)$...otherwise it wouldn't be the greatest lower bound. So it's a limit point because it's not in the set, but there are infinitely many elements in $S$ that are close to $u$. Does that work? Apr 24, 2017 at 12:45
• $\inf S$ can be in the set (for example if $S$ is closed, i.e. $S=\bar S$) Apr 24, 2017 at 12:50

You should replace the following statement in the third statement:

\begin{align} \text{For every} \; \varepsilon > 0, \text{there exists an} \; a \in S \; \; \text{such that} \; \;u \geq a > u + \varepsilon.\end{align}

Now choose $\varepsilon = 1/n$ inductively and create the desired sequence, as suggested by Pawel.

• Hmmm I'm not sure how choosing $\varepsilon = \frac1n$ helps because as you wrote above, isn't it for every $\varepsilon$? Apr 24, 2017 at 12:56
• @AR7 That's exactly the point; since the statement holds for any $\varepsilon$, you are free to choose a value of $\varepsilon$, and still ensure the resulting condition holds. Choosing $\varepsilon$ as mentioned above allows you to pick countable many values of $\varepsilon$, and a countable number of $a's$ satisfying the condition above. You can easily check that this construction allows you to construct a sequence that converges to $u$, guaranteeing that $u \in \bar{S}$. Apr 24, 2017 at 12:59
• Ah I see now...last question: Once I've picked $\varepsilon = \frac1n$, what is "the desired sequence" exactly? Is it something like saying $\varepsilon \rightarrow 0 \Rightarrow u + \varepsilon \rightarrow u \Rightarrow u \ge a \gt u \Rightarrow a = u$ Apr 24, 2017 at 13:04
• @AR7 Kind of, since your statement implies than the sequence of values of $a's$ picked will converge to $u$, as desired. But what are those values of $a'$? Well, we don't need to know them explicitly; rather, we simply need to guarantee their existence. Pick $\varepsilon = 1$; using the condition mentioned above, you can pick $a_1$ as desired. Now pick $\varepsilon = 1/2$; using the condition mentioned above, you can pick $a_2$ as desired. Now apply the definition inductively and convince yourself of the argument that is being used. Apr 24, 2017 at 13:10
• @AR7 Also, note that this also proves the statement that if $u = \inf(S)$, then there exists a sequence in $S$ such that $x_n \longrightarrow S$. If we we were initially given a sequence such that $\{ x_n \} \subset S$, then this also shows that corresponding to this sequence, there exists a subsequence that converges to $u$, which is an interesting result in its own right. The initial sequence may not have been monotonically decreasing; if it were, then it would have converged to $u$. Nevertheless, the existence of a subsequence is guaranteed, as shown. Apr 24, 2017 at 13:13

If $u$ is the infimum of $S$, then for every $\delta > 0$ there is some $x \in S$ such that $u \leq x < u +\delta$; so $u$ lies in the closure of $S$.