# Compactness, why is $(0,1)$ not compact? I need the “thought process” [duplicate]

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.

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.

## marked as duplicate by JMoravitz, Takumi Murayama, Captain Lama, John B, AretinoApr 28 '16 at 21:17

• Your cover consists of sets of the form $(\epsilon,1)$ where $\epsilon >0$. When you say "$\epsilon \to 0$ alone would cover $(0,1)$", note that this limit (in the sense you want to define it) is not part of your cover. – s.harp Apr 28 '16 at 18:49
• A finite subcover of $(\epsilon,1)_{\epsilon>0}$ has a smallest $\epsilon$. Suppose it is $\epsilon_0$, then the subcover fails to cover the point $\frac{\epsilon_0}{2}$. – almagest Apr 28 '16 at 18:50
• Hi, so your points are that $(\epsilon,1)$ in fact does NOT (open) cover $(0,1)$. Then, why should I even consider $(\epsilon, 1)$ then? If I wish to show $(0,1)$ is not compact I must 1. Find an open cover of $(0,1)$ 2. Show that the open cover in (1.) does not have a finite subcover, yes? But $(\epsilon,1)$ doesn't even satisfy (1.) as you say it doesn't cover $(0,1)$ to start with so it doesn't qualify to show $(0,1)$ in non-compact... I am very well confused now. – John Trail Apr 28 '16 at 18:55
• The "interval" $(\epsilon,1)_{\epsilon > 0}$ does not refer to a single interval. It refers to an infinite set of intervals, for varying $\epsilon$, where the set of those $\epsilon$ is all numbers between 0 and 1. – Mees de Vries Apr 28 '16 at 18:57

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).

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.

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.

• @John Trail If you look at this and go "huh, $\cup_{0<x<1}\{(x,1)\}$ looks like $\lim_{x\to 0} (x,1)$" then yore doing something very right. – Stella Biderman Apr 28 '16 at 19:18
• This is really messing with my mind...My question is then when you say $C=\{(\frac{1}{2},1),(\frac{1}{4},1),(\frac{1}{8},1)...\}$ covers every point in $(0,1)$, how about then I keep eliminating the obvious overlaps e.g. $(\frac{1}{2},1)$ which is clearly covered by the remaining open intervals $(\frac{1}{4},1),(\frac{1}{8},1)..$ and do the same for $(\frac{1}{4},1)$...which will ultimately leave me with some $(\epsilon,1)$? If someone argues $\frac{\epsilon}{2}$ isn't covered! I say sure then I can keep repeating $(\frac{\epsilon}{2},1)$ and get rid of $(\epsilon,1)$. This seems finite... – John Trail Apr 29 '16 at 13:37
• So essentially, I can keep iterating this..."create a new open interval which covers the points not covered by a previous cover $(\epsilon,1)$" for every point in $(0,1)$ and then "eliminate the redundant(overlapping) open covers" to constantly keep me with a finite (namely, well, just one open cover) for $(0,1)$? – John Trail Apr 29 '16 at 13:39
• @JohnTrail You cannot, in fact, "eliminate the redundancies". If you get rid of finitely many intervals from this cover, it remains a cover; but if you get rid of infinitely many intervals, it might not be a cover anymore. Informal arguments like "essentially, I can keep iterating this" are dangerous; can you explicitly write down a finite subcover of this cover? Keep in mind you can't use "$\epsilon$", it's just a variable. – Noah Schweber Apr 29 '16 at 17:06

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 $(0,1)\subset \bigcup_{i=1}^n(\epsilon_i,1)$.

• Hi, but this here " $(0,1)\subset \bigcup_{\epsilon>0}(\epsilon,1)$ (in fact it is equal)" I don't understand. The comments above have pointed out that say $\epsilon_0$ is the smallest element of $\epsilon$ then it fails to cover $\frac{\epsilon_0}{2}$. Doesn't this mean there are points left out (i,e, failed to cover) in $(0,1)$? – John Trail Apr 28 '16 at 18:59
• This is the definition of a cover. Every point of $(0,1)$ lies in some $(\epsilon,1)$. But if you have just a finite number of $\epsilon$s, some points will be left out yes, so it won't be a cover. – Augustin Apr 28 '16 at 19:02
• @JohnTrail It might help if you try to come up with a concrete example of a point left out of the union. For example, is ${1\over 2}\in \bigcup_{\epsilon>0}(\epsilon, 1)$? Why? (Once you understand this, I think the general statement will be clearer.) EDIT: Also, what do you mean by "smallest element of $\epsilon$?" – Noah Schweber Apr 28 '16 at 19:09
• @NoahSchweber So clearly if say, $\epsilon=1/3$ for one of the $epsilon$ in the union, then $1/2$ IS in the union...i.e. covered by the union. The "smallest" $\epsilon$ probably refers to me claiming that the open cover of $(0,1)$ by $(\epsilon,1)$ is FINITE. i.e a minimal $\epsilon$ exists. The counter argument I was given was that, "well how abut $\epsilon/2$? Or $\epsilon/3$? Those aren't covered." But then my counter argument to that is now "then I can "renew" or "improve" the cover to $(\epsilon/2,1)$ and get rid of $(\epsilon,1)$. Similarly, I can keep going, – John Trail Apr 29 '16 at 13:49
• @JohnTrail It does not. It might be easier to think about how there is no largest natural number. You say: "But what about $N$?" I say: "$N$ isn't the largest, since $N+1$" is bigger." You say: "OK, well replace $N$ by $N+1$, now that's the largest." I say: "No, $N+2$ is still bigger." You say: "I can keep going . . ." But this ability to keep going actually means that you are wrong: in order for there to be a largest natural number, you need to pick one and stick with it. Similarly in this context, you can't keep changing your interval if you think there's a biggest one. – Noah Schweber Apr 29 '16 at 17:08

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: $(\epsilon_1,1)\cup\cdots\cup(\epsilon_n,1)=(\min(\epsilon_1,\ldots,\epsilon_n),1)$.