0
$\begingroup$

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 $(0,1)$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $x < 1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

$E_x := (0,x)$ 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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.