TITLE: Infimum of a set is its limit point
QUESTION [5 upvotes]: Lemma: $A\subset\mathbb{R}, z=\inf(A)$, and $z \notin A$, then $z$ is an accumulation point of $A$.
Here is my proof: Let $e>0$ be given. Consider $N(z,e)$.  Note that $N(z,e)=(z-e,z+e)$.
Since $z=\inf(A)$, $z+e$ cannot be a lower bound of set $A$.  Hence there exists a $t \in A$, such that $z-e<t<z+e$, therefore $N(z,e)$ intersects with $A$, which is non-empty.
$N$ represents the Neighborhood of $z$.
Any feedback or corrections would be very helpful thank you.

REPLY [3 votes]: Two relatively minor points:you get $a \in A$ where $z \le a <z+e$, as $z$ is already a lower bound.  Where you use that $z \notin A$ is the fact that $z < a$ must hold (otherwise we would have that $z = \min(A)$ and this could be isolated: $\{0\} \cup[1,2]$ is an example where this happens). So we have $a \in (A\setminus \{z\})\cap N$ for every neighbourhood $N$ of $z$. This makes it a limit point in my terminology (and as the reals are $T_1$, it is also an ($\omega$-)accumulation point: every $N$ intersects $A$ in an infinite set.