In my class the word "cover" is used very informally such as this set covers another set (this is for a class in PDE not topology by the way).
Can someone provide a trivial example of cover to get the point across?
Say if I want a set to cover the real line, what do I mean by set? Can I cover a line segment? Can the set be points?
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$\begingroup$Take some space $X$. A cover for $X$ is a bunch of sets, say $U_1, U_2, ...$ so that $X \subseteq U_1 \cup U_2 \cup ...$ (That is, $X$ is contained in the union of $U_1, U_2, ...$)
In layperson terms, a cover for a set $X$ is a bunch of sets so that $X$ is completely contained in that bunch of sets.
Other cool things:
We could also define an open cover. An open cover is the exact same thing as a cover, except each $U_1, U_2, ...$ is open.
We could also define a finite cover. A finite cover is simply a cover $U_1, U_2, ..., U_n$ with a finite number of $U_i$'s.
We could also define a subcover. A subcover is simply a cover that is contained within another cover. So say $U_1, U_2, U_3$ covers $X$. But let's say, it happens to be the case that, hey, $U_1, U_2$ also covers $X$. Well, then $U_1, U_2$ is a subcover of $U_1, U_2, U_3$. They are both perfectly valid covers--it just so happens that one is contained in the other.
The idea of a cover gives rise to compactness, which is can be a very powerful mathematical concept. A space $X$ is compact if, for every open cover (remember, an open cover is a bunch of sets that are each open, so that $X$ is contained in that bunch of sets), we can find a finite subcover (so we can find a cover within that open cover that is finite). So, for every bunch of open sets that you can find that covers $X$...if I can take that bunch of open sets, dwindle it down to all but a finite bunch of open sets that still covers $X$, then $X$ is compact.
$\endgroup$ 6 $\begingroup$Definition. $X$ be a topological space, $A \subseteq X$ a subspace (not necessarily a strict one). An open cover of $A$ in $X$ is a collection of open sets $\{\mathcal{U}_i\}$ in $X$ such that $A \subseteq \bigcup_i \mathcal{U}_i$.
Geometric meaning. An open cover of $A$ is a collection of (possibly overlapping) open sets in $X$ which, after looking at their union globally, contains $A$ inside. This where the terminology "cover" comes from : you're using small, simple open sets to fill up a rather complicated-to-inspect subspace $A$ of $X$.
As an example, consider the circle $S^1 = \{z : |z| = 1\}$ embedded in $\Bbb C$. Consider the collection of open sets $\{\mathcal{U}_i\}$ where $\mathcal{U}_i$ are "thickened" arcs of the circle $S^1$. This covers $S^1$.
Importance. Open covers are not necessarily always "nice". For example, consider the open cover $\{(1/n, 1)_{n \in \Bbb N}\}$ of the interval $(0, 1)$. So, you might think that it's just a technical term and of no particular use. However, you're just missing the point.
Open covers are a language to shift from the topological point-of-view to a combinatoric point-of-view. More precisely, instead of inspecting properties of a topological space $X$ by hand, one tries to inspect combinatorial properties of the open covers of $X$, which sometimes turns out to be a very easy thing to do due to having less structure.
For example, consider $(0, 1) \subset \Bbb R^2$. This is most certainly not homeomorphic to, say, $[0, 1]$. An obvious argument is that the former is something "large" (homeomorphic to $\Bbb R$), while the latter is "small". The correct analytic formalization of this idea is sequential compactness, i.e., that every infinite sequence has a convergent subsequence. One can use some analysis to show that $[0, 1]$ is sequentially compact while $(0, 1)$ is not. However,
$\text{Fact}$ : In good spaces, sequential compactness is equivalent to the property "every countable cover has a finite subcover".
A complete characterization of good spaces is a bit complicated, but metric spaces are examples of such spaces. In any case, note that the open cover of $(0, 1)$ produced in the answer above has no finite subcover, while every open cover of $[0, 1]$ has a finite subcover (try proving this!). Thus, we have just used a combinatorial property of open covers to get back analytic properties of the total space. In particular, the property of open covers I have mentioned above is called (countable) compactness of topological spaces.
Another cool thing about open covers is the Lebesgue covering lemma, which states that given a compact metric space with an open cover, there is a positive real number $\epsilon$ such that each open set of diameter at most $\epsilon$ fits inside an open set in the cover. This is a very nice theorem, which in turn is used to prove the fact I have mentioned above. This also has interesting applications in algebraic topology (for example, in the proof of excision theorem for singular homology).
$\endgroup$ $\begingroup$For the interval $[0,6]$ the set of $\{B_1(0),B_1(1),B_1(2),B_1(3),B_1(4),B_1(5),B_1(6)\}$ is a cover, where $B_\epsilon(x)$ is the open ball of radius $\epsilon$.
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