Justification that (a,b] in the standard topology is not closed? I was wondering how to justify that $(a,b]$ is not closed in the standard topology on R. I know that the definition of a closed set is one whose complement is open. So then we have that 
R - (a,b] = $(-\infty,a] \cup (b,\infty)$. 
So the set $(b,\infty)$ is an open ray which is open in the standard topology. And the set $(\infty,a]$ is closed because its complement is the open ray $(a,\infty)$.
So I have that R - (a,b] is the union of a closed set and an open set but does that necessarily mean then that $(a,b]$ is not closed? It doesn't seem like quite enough to me. Perhaps there is some way to write $(\infty,a]$ as an infinite union of basis elements in the order topology that I just haven't thought of? 
P.S. I was hoping to get clarification from as pure of a topological standpoint as possible in terms of sets in the topology or basis for the topology, I know there are lots of other ways to answer this question if you look at sequences, seq. compactness and the metric on R 
 A: In the standard topology in $\mathbb{R}$, a set $U$ is open if and only if for every $x\in U$ there is an interval $(c,d)$ such that $x\in (c,d)\subseteq U$. Take $x=b$. If $x\in (c,d)$ then we have $c<b<d$, and so $(c,d)\not\subseteq (a,b]$.
A: Note that a set $V \subseteq \Bbb R$ is open if and only if for all $v \in V$ there exists some $\varepsilon >0$ such that the interval $(v-\varepsilon, v+\varepsilon) \in V$.
If we consider the set $(-\infty,a]\cup(b,\infty)$, we find that for the point $a$ we run into trouble, since there does not exist a $\varepsilon$ such that the above definition of open sets holds. Therefore the set is not open and subsequently, $(a,b]$ is not closed.
A: (This had been a comment, but I thought it might be of sufficient interest to others to post as an answer, even though it doesn't really answer the spirit of the question.)
For a hammer proof (see here also), the Tietze extension theorem says that for a normal topological space, every continuous function from a closed set in the space to the reals has a continuous extension to the whole space, but $f: (0,1] \rightarrow {\mathbb R}$ defined by $f(x) = \sin \frac{1}{x}$ is continuous but has no continuous extension to all of ${\mathbb R},$ so $(0,1]$ can't be a closed set.
