I'm having trouble picturing what compact sets and open sets actually are. Open and closed intervals make enough sense to me, but for whatever reason, moving to the next level of abstraction is difficult for me.
I've read the definition in my textbook (Ross), but it didn't help as much as I would like.
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In the beginning it is always hard to understand these concepts, since they seem to lack the motivation. Don't worry you'll get the hang of it. The best way to learn mathematics is by doing it. Don't just sit around and think about these concepts, try to apply them. Try and solve a problem on your own, doesn't matter if it takes weeks, that's the only way you'll ever learn.
I found these videos on Youtube, which I think are quite useful. This one is particularly on Compact Sets. The whole lecture series is amazing. I think he talks about open sets at around Lecture 8 or 9.
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The open subsets of the line do include the open intervals, but they also include sets that are the union of two open intervals, three open intervals, or even the union of an arbitrary collection of disjoint open intervals.
e.g. the set of all real numbers that are not integers is an open set.
The open intervals form a basis for the topology: you can detect whether or not a set $S$ is open by checking, for every point $x$, that there is an open interval $I$ such that $x \in I \subseteq S$.
Many topologies can be expressed in terms of a basis: in the plane, a common choice of basis is to let the basis open sets be the interiors of circles. Note that this doesn't mean all open sets look like circles: e.g. the set of points inside of a square forms an open set (because every point is contained in a circle that is contained entirely within the square).
If you're algebraically inclined, it's worth noting that if $f$ is a continuous real-valued function on a topological space, then the solution set to an inequation $a < f(x) < b$ always forms an open set. Also, the solution set to an inequation $a \leq f(x) \leq b$ always forms a closed set. ($a$ and $b$ are allowed to be $\pm \infty$)
Compact sets tell us something in the opposite direction: if $S$ is a compact set and $f$ is a continuous real-valued function, then $f(S)$ is also a compact set. (in the real line, the compact sets are the sets that are closed and bounded)
Properties of a topological space such as Hausdorff, regular, normal, etc. require that there be lots of open sets in the space. Compactness requires that there not be too many open sets in the space. (Example: with the discrete topology, where every point is an open set, an infinite space can't be compact.) These properties "fight" each other. So a compact Hausdorff space has a delicate balance of not too few and not too many open sets.