Show that infimum of set is in closure of set.

Question

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?

Answer 1

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.

Answer 2

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 $S$.