For $E_x := (0,x)$ where $0<x<1$, is $\epsilon := \{E_x:0<x<1\}$ an open cover of $(0,1)$?

We can prove that each $E_x$ is open; take $y\in E_x$ and let $r = \min\{d(0,y), d(x,y)\}$. Then $N_r(y)\subset E_x$. Since this is for any $y\in E_x$, then $E_x$ is open.

If it is the case that $\epsilon = \{E_x\}$ is an open cover, how do we prove this fact?

Moreover, suppose $\epsilon$ is an open cover. We want to show it has no finite subcover of $(0,1)$. Is the following proof correct?

Suppose there exists $x_1, x_2, ..., x_n$ such that $\bigcup_{j:1\leq j \leq n}^{n}E_{x_j}\supset(0,1)$. Choose $x = \max_{1\leq j \leq n}(x_j)$. Then $E_x = (0,x) \subset (0,1)$. Since all other $E_x$ are contained in this $E_x$, the union of the $E_x$'s cannot be a finite subcover of $(0,1)$.

  • $\begingroup$ The notation is a little misleading: $E_x$ is a set of intervals, and this set does not depend on $x$. It would be clearer in my view to set $E_x := (0, x)$ and define the (candidate) open cover to be $\mathcal{E} := \{E_x : 0 < x < 1\}$ . $\endgroup$ – Travis Willse Oct 24 '15 at 21:52
  • $\begingroup$ thanks. i fixed the notation $\endgroup$ – socrates Oct 24 '15 at 21:58
  • $\begingroup$ Actually, all in all there is no need for the E_x notation at all. It's not incorrect. just not needed. $\endgroup$ – fleablood Oct 24 '15 at 22:22

To show that $\epsilon$ is an open cover, it's enough to show (1) that its elements are all open (which has already been done in the question statement) and (2) that its union is $(0, 1)$.

To show (2), it's enough for each $y \in (0, 1)$ to show that there is some $E_x \in \epsilon$ such that $y \in E_x = (0, x)$. Can we find such an $x$?

The proof that the cover has no finite subcover (and hence that $(0, 1)$ is noncompact) is almost correct: One needs to show that the union $E_x = (0, x)$ is not all of $(0, 1)$, but this is immediate, as $x < 1$.

  • $\begingroup$ Such an $x$ would simply be any $y<x<1$, correct? i.e: Take some $x\in (0,1)$. There is a $y$ s.t. $0<x<y<1$. Then $E_x\subset E_y$ and $E_y\subset (0,1)$. We then have $\bigcup_{x}{E_x} = (0,1)$. So we know $\epsilon$ is an open cover of $(0,1)$. (apologies for swapping x, y; i copied this over) $\endgroup$ – socrates Oct 24 '15 at 22:53

$E_x := (0,x)$ is $\epsilon := \{E_x : 0<x<1\}$ an open cover of (0, 1)?

Is it an open cover?

1) Are the $E_x$ each open? Yes. You showed that. Also presumably earlier in the course you were shown all open intervals are open sets.

2)Do they cover (0, 1)? Yes, if $x \in (0, 1)$ then $0 < x < 1$ and we can find a y such that $0 < x < y < 1$ so $x \in E_y$ and $E_y \in \epsilon$. So, yes $\epsilon$ covers (0, 1)

So $\epsilon$ is an open cover.

Does $\epsilon $ have a finite subcover.

You argued correctly that if so then there would be an $E_y = (0,y) \subset (0,1)$ for all other of the $E_x$ are subsets of $E_y$ so $E_y$ would have to cover (0,1). BUT you didn't show it doesn't. Because you didn't show $E_y$ was a proper subset.

It is and it doesn't.

It doesn't because because $y \notin E_y$. so $E_y$ doesn't cover (0,1). So there is not finite subcover.

  • $\begingroup$ @Did Ooooooooooops! $\endgroup$ – fleablood Oct 24 '15 at 22:19
  • $\begingroup$ For 2), the argument obviously makes sense. However, is this as rigorous as needed to move from the $E_y \in \epsilon$ to $\epsilon$ covering (0,1)? I guess I am struggling to grasp the idea of an open cover, and perhaps a more rigorous justification of the intuitive idea that you can 'keep adding on ys ' to eventually get the complete union $\endgroup$ – socrates Oct 24 '15 at 22:29
  • $\begingroup$ I think it is. A cover is a colllection of sets that "cover" a set. In other words $E = \{S_\alpha|$ a bunch of sets$\}$ "covers" $A$ if $A \subset \cup_{S_\alpha \in E}S_\alpha$. In other words: for all $x \in A$ then $x \in S_{\alpha}$ for some $S_{\alpha} \in E$. That's all. $\endgroup$ – fleablood Oct 24 '15 at 22:41

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