Compactness, why is $(0,1)$ not compact? I need the "thought process"

Question

See, I am told that $(0,1)$ is not compact as a subspace of $\mathbb{R}$. Question is, how do I conclude that?

The hint says

$(\epsilon,1)_{\epsilon>0}$ does not have a finite subcover.

But I am unsure why $(\epsilon,1)$ does not have a finite (open) subcover for $(0,1)$. The right end point is fixed to $1$ so all I can do is move around $\epsilon$. So some $\epsilon$ as close as possible to zero $(\epsilon,1)$ where $\epsilon \to 0$ alone would be an open cover of $(0,1)$.

Any other $\epsilon>0$ I choose would be redundant. $(1/2,1),(1/60,1),(8/7,1)...$ whatever. These are already covered by $(\epsilon,1)$ where $\epsilon \to 1$. So that means, the open cover by $(\epsilon,1)$ is finite on its own to start with. namely, just one open subset $(\epsilon,1)$ of $\mathbb{R}$ covers $(0,1)$.

I can come up with loads of other open covers, sure, but as I said, any other open cover apart from $\epsilon \to 0$ can be removed and still be an open cover of $(0,1)$. e.g. $\{(\epsilon,1)_{\epsilon \to 0},(23/81,1)\}$ is an open cover of $(0,1)$ but I can get rid of $(23/81)$ if I want to since that still leaves me with an open cover of $(0,1)$. So the same for any infinite open cover $\{(\epsilon,1)_{\epsilon \to 1},(1/2,1),(2/5,1),(1/3,1),...\}$ I can get rid of everything that follows $(\epsilon,1)$ and be left with an open cover i.e. finite subcover.

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I need some clarification here and most probably; I am mistaken with something so blatantly obvious to those with experience. Think I am a super stupid human being and explain from possible the $1+1=2$s of this bizarre world of compactness if necessary. I think someone needs to fix some basic but esoteric notion for this one.

Answer 1

You write

some $\epsilon$ as close as possible to zero

But there is no such $\epsilon$! For any positive $\epsilon$, there is a strictly smaller positive $\epsilon$ - for instance, $\epsilon/2$. And the interval $({\epsilon/2}, 1)$ will contain some points in $(0, 1)$ that $(\epsilon, 1)$ does not (for instance, $\epsilon$ itself).

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While no specific interval $(\epsilon, 1)$ covers $(0, 1)$ (for $\epsilon>0$), the point is that a cover is a collection of intervals; to be a bit more concrete, consider the collection of intervals $$\mathcal{C}=\{({1\over 2}, 1), ({1\over 4}, 1), ({1\over 8}, 1), . . . \}.$$ Every point in $(0, 1)$ is in one of these intervals, so $\mathcal{C}$ is a cover of $(0, 1)$.

The specific cover being talked about above is $$\{(\epsilon, 1): 0<\epsilon<1\};$$ again, an infinite collection of intervals.

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EDIT: It seems like there might also be some confusion about what infinite unions mean. Remember that $\bigcup A_i$ is the set of all things which are in some $A_i$: $$\bigcup_{i\in I}A_i=\{x: \exists i\in I(x\in A_i)\}.$$ So if $x\in A_0$ but $x\not\in A_1$, then $x\in \bigcup A_i$: the fact that $x\not\in A_1$ doesn't matter, as long as $x$ is in some $A_i$ (in this case, $A_0$) it's in the union.

Answer 2

First $(\epsilon,1)_{\epsilon>0}$ is an open cover of $(0,1)$ because:

• for any $\epsilon>0$, $(\epsilon,1)$ is open
• $(0,1)\subset \bigcup_{\epsilon>0}(\epsilon,1)$ (in fact it is equal)

Now assume that this open cover has a finite subcover. This means that there exists $\epsilon_1,\dots,\epsilon_n$ such that $(0,1)\subset \bigcup_{i=1}^n(\epsilon_i,1)$. Now let $\epsilon_0=\min(\epsilon_1,\dots,\epsilon_n)$. As $\epsilon_0>0$, there exists $x\in (0,1)$ such that $x\in(0,\epsilon_0)$. So $x$ belongs to none of the intervals $(\epsilon_i,1)$, wich contradicts the fact that $(0,1)\subset \bigcup_{i=1}^n(\epsilon_i,1)$.

Answer 3

I think you are confused with the definition of cover and compactness. Let's deal only with subsets of $\mathbb{R}$ to make things clearer:

Definition: A cover of a subset $A$ of $\mathbb{R}$ is a collection $\mathcal{C}$ of subsets of $\mathbb{R}$ such that $A\subseteq \bigcup\left\{B:B\in\mathcal{C}\right\}$, that is, the union of all elements of $\mathcal{C}$ contains $A$.

Note that $\mathcal{C}$ above is a collection of sets (or a "set of sets", although this kind of nomenclature is usually avoided), so it is ''one level above $A$''. We can express this as: $A$ is an element of the power set of $\mathbb{R}$, whereas $\mathcal{C}$ is a subset of the power set of $\mathbb{R}$: $$A\in\mathcal{P}(\mathbb{R}),\qquad\mathcal{C}\subseteq\mathcal{P}(\mathbb{R}),\qquad\text{so }\mathcal{C}\in\mathcal{P}(\mathcal{P}(\mathbb{R})).$$

Definition: We say a cover $\mathcal{C}$ of $A$ is open if the elements of $\mathcal{C}$ are open subseteq of $\mathbb{R}$. We say that $\mathcal{C}$ is finite if $\mathcal{C}$ is a finite set. A subcover of $\mathcal{C}$ is a subset $\mathcal{D}\subseteq\mathcal{C}$ which is itself a cover of $A$. Finally, $A$ is said to be compact if every open cover $\mathcal{C}$ of $\mathcal{A}$ admits a finite subcover.

Let's give an examples:

Example: Let $F=\{a_1,\ldots,a_n\}$ be a finite subset of $\mathbb{R}$. Let's show that $F$ is compact: Let $\mathcal{C}$ be any open cover of $F$. We need to find a finite subcover of $\mathcal{C}$. By hypothesis, $$\{a_1,\ldots,a_n\}=F\subseteq\bigcup\{C:C\in\mathcal{C}\}$$ so for each $i$, there is some $C_i\in \mathcal{C}$ with $a_i\in C_i$. Fix such $C_i$ and define the $\mathcal{D}=\left\{C_1,\ldots,C_n\right\}$. Then $\mathcal{D}$ is a subset of $\mathcal{C}$ for which $F\subseteq\bigcup\left\{C_i:C_i\in\mathcal{D}\right\}$, so it is a subcover, and it is clearly finite. (We did not use the hypothesis that the elements of $\mathcal{C}$ are open, but this doesn't matter in this case.)

Your example Let $A=(0,1)$. In order to show that $A$ is not compact, we need to contradict the definition of compactness. This means that we must prove that there exists an open cover $\mathcal{C}$ of $(0,1)$ which does not admit subcovers. The hint is to define $\mathcal{C}=\left\{(\epsilon,1):0<\epsilon<1\right\}$. This is a cover of $(0,1)$ (verify this: given $t\in(0,1)$, find some $\epsilon$ for which $t\in(\epsilon,1)$.) and its elements are open, so we are half-done. The last part is to verify that $\mathcal{C}$ does not admit a finite subcover. A finite subcover has the form $\left\{(\epsilon_1,1),(\epsilon_2,1),\ldots,(\epsilon_n,1)\right\}$ for certain $\epsilon_1,\ldots,\epsilon_n\in(0,1)$. So you simply need to find $t\in(0,1)$ for which $t\not\in(\epsilon_1,1)\cup\cdots\cup(\epsilon_n,1)$. I'll leave that to you (Hint: $(\epsilon_1,1)\cup\cdots\cup(\epsilon_n,1)=(\min(\epsilon_1,\ldots,\epsilon_n),1)$.