Is the interval $(-\infty,0]$ open, closed, or neither? Consider the set
\begin{equation}
A=(-\infty,0]
\end{equation}
Is this set open, closed, or neither?

Let's check. First, note that its complement, $A^\mathsf{c}=(0,\infty)$, is open, which implies that $A$ is closed. However, $A$ appears to be open on the left. I am a bit embarrassed to say I'm not sure whether it is closed or half-open.
Thanks!
 A: it is closed as a set (though neither, as an interval). A closed set does not mean the same as a closed interval: The latter simply means that the interval contains both endpoints.  
For that matter as EthanAlvaree pointed out in a comment, the answer may depend on the topological space under consideration. It seems it is the reals (so then my answer above applies), indeed if the complement of $A$ is $(0,\infty)$ as you stated then it is understood that the whole space is $A$ union its complement, that is $(-\infty,0]\cup(0,\infty)=(-\infty,\infty)$. 
But, if the space is a compactification of the reals e.g. something like $[-\infty,\infty]$ then the set $A$ is not closed. 
On the other hand if the whole space is $A$ itself, then $A$ is both open and closed. (Alternatively, this also happens if the whole space is the reals, but with the discrete topology.) 
Well, yes if you do not make a difference between a closed set and a closed interval (except that a set need not be an interval), then this would be indeed confusing. But I think the accepted terminology regarding intervals is more specific than treating them as simply sets in a topological space. You could talk about intervals any time you have a linear order, even if you have no topology, although you could use intervals to define the so-called order topology ... but you may consider an entirely different topology on your ordered set, and that would not affect the meaning of the terms open, closed, and half-open, as applied to intervals. 
A: It's closed. Your check is correct. There is potentially conflicting terminology with the words open and closed: on the one hand, the words make sense for parts in any topological space. On the other hand, they can be used to qualify the bounds of an interval. At $\pm \infty$, the open symbol is always used, it's not relevant. So the status of the interval depends only on what happens at the other bound. Note that the interval $(-\infty, +\infty)$ is of course open and closed.
A: To determine if $A$ is open, we need to examine if every $x \in A$ has some $\varepsilon > 0$ neighborhood around $x$ completely contained in $A$. To this end, think about any $\varepsilon$ neighborhood centered around $x=0$, and you can deduce whether $A$ is open from there.
A: It is closed as the preimage under a continous function of the closed set $[0,1]$
$$
(e^x)^{-1}([0,1])
$$
