Covers, and compact sets Please go to this lecture(1). I have numbered my questions for organization and added the picture which I will constantly refer to. I also apologize in advance for this lengthy question


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*Right about here, he says that the point (the top one "enclosed" by the semi-circle) isn't open in $X$. I don't understand this. Why exactly isn't that open in $X$

*Also he later talks about the semi-circle neighborhood being in $Y$ and that makes the point on the boundary an interior point. This confuses me. He left out the other part of the circle and just ignored that the part of the real "ball" lies outside $Y$ and therefore that ball isn't contained in $Y$.

*Also are the sets $X$ and $Y$ even open sets in the lecture? Because later in the follow up video he writes $E = Y \cap G$. How can a metric space (is it even a set?) intersect with a set? Aren't they two different things? Isn't like saying $5 \cap [0,1]$? Sorry if the question is ignorant here. 

*Later in here, he tries to prove that $E = Y \cap G \iff E$ is open. He says that if a neighbourhood is contained in G, then the intersection of that neighbourhood and $Y$ is a neighbourhood in both $Y$ and $E$. But what happens if he takes a point that's not even in $Y$? And draw a neighbourhood around it such that it doesn't even intersect $Y$ or $E$? And suppose $Y$ really is a set (assuming (3) had been answered and clarifed), How does that imply this mysterious set intersection with an open set gives me and open set? That can't be true in general. What if $(-2,2)\cap [-1,1] = [-1,1]$? That doesn't give me an open set

*Finally, I don't understand the idea and strategy in the forward direction of the proof. Like at all...

*I found this proof (part of (5))  , in one line they say that $d(p,q) < r_p, q \in Y \implies q \in E$. How does that make sense? What if $q$ is outside of $E$? 

 A: *

*He's saying that the semi-circular set is open in $Y$ but not in $X$.  He's not talking about the point itself.  The reason this set isn't open in $X$ is because points along the top edge of the rectangle $Y$ have the property that every open ball contains a point in the set $Y$ and a point not in the set $Y$ (draw a little circle around a point lying on that edge, you'll see what I mean). 

*I'm not entirely sure what you mean.  Could you like to a time in the video?

*$X$ is a metric space to start with, which is a set "endowed" with a metric.  Then $Y$ starts out life as an open set in $X$, which we then 'turn into' a metric space by giving it the restricted metric.  As sets, $Y\subset X$.  For $E=Y\cap G$, you think of this intersection as plain old ordinary set intersection - $Y$ is still a set even though we've given it a metric. 

*(and 5.) It's definitely important to note that he's showing $E$ is open in $Y$.  For the forward direction, we're actually constructing the set $G$ so that (a) it is open in $X$ and (b) $E=Y\cap G$.  You simply construct this by drawing small enough neighborhoods around all the points in $E$.  These neighborhoods might 'spill over' into $X$, but that's okay.  Then, since neighborhoods are open, we can take an arbitrary union of them (in this case, take the union over all points in $E$) and the result will be an open set.  You can then check that (a) and (b) hold.   For the reverse direction, it doesn't matter if he picks a point that isn't in $Y$.  All we need to show is that each point that is in $E$ has a neighborhood that is contained in $E$.  $x\in E$ means that $x\in G$, and $G$ is open so we have our neighborhood.  Like he says, this neighborhood  intersected with $Y$ is a neighborhood in $Y$, which is what we need.

*That such an $r_p$ exists is guaranteed by the assumption that $E$ is open relative to $Y$.  I.e. $q$ can't be outside of $E$ because I've said that $d(p,q)$ is small enough so that it isn't.  Try drawing a picture.
