Why can't a open interval in $\mathbb{R}$ be compact?

Question

If I choose $(0,1)$ and do a covering like this:

$(-1,1/2)$ and $(1/3,2)$ it's a finite covering, then $(0,1)$ is a compact set?

(1) It's not a covering -- 1/2 isn't covered -- and (2) EVERY open covering of the set must contain a finite sub covering. — Christopher Carl Heckman, Apr 25 '16 at 07:40
every open cover must have a finite subcover. The classic example would be {(1/n, 1 - 1/n)} is an open cover which has no finite subcover. — fleablood, Apr 25 '16 at 07:43
So the set of answers is compact, now im so happy I can't chose one. — José Osorio, Apr 25 '16 at 07:53
Similar questions from the past: http://math.stackexchange.com/questions/257514/cover-of-0-1-with-no-finite-subcover-open-sets-of-compact-function-spaces http://math.stackexchange.com/questions/609217/why-is-an-open-interval-not-a-compact-set http://math.stackexchange.com/questions/1133580/an-example-of-an-infinite-open-cover-of-the-interval-0-1-that-has-no-finite-su http://math.stackexchange.com/questions/1245499/proof-of-0-1-is-not-compact-with-usual-metric — Martin Sleziak, Apr 25 '16 at 08:44

score 7 · Answer 1 · answered Apr 25 '16 at 07:48

If I choose $(0,1)$ and do a covering like this: $(-1,\frac{1}{2})$ and $(\frac{1}{3},2)$ it's a finite covering, then $(0,1)$ is a compact set?

You are misunderstanding the definition of a compact set. The definition starts with "For every open coverings, there exists a sub coverings......". So, in order to show that $(0,1)$ is compact, you first have to prove something that is true for every open coverings of $(0,1)$, which is quite hard.

The example you gave $(-1,\frac{1}{2})\cup(\frac{1}{3},2)$ is indeed an open covering. However, this does not help you much since you will need every possible open coverings.

In fact, to show a set is not compact, all you need is just one counterexample which is already given in other answers. To show a set is compact, however, is much more difficult. Sometimes you can use the definition of compactness directly but as I said, doing something to every possible open covering sounds difficult. In these case, people usually either argue by contradiction or use the fact that being closed and bounded sometimes implies compactness.

score 3 · Accepted Answer · answered Apr 25 '16 at 07:41

3

$(0,1)$ is not compact.

Proof: Consider the open covering $\displaystyle\left\{\left ({1\over n}, 1 - {1\over n}\right) \right\}_{n=3}^\infty$; this is an open covering of $(0,1)$ but it has no finite subcovering.

answered Apr 25 '16 at 07:41

Christopher Carl Heckman

4,458

User001 · Answer 3 · 2016-04-25T07:45:59.137

2

Take the sequence $\large a_n = \frac{1}{n}$, which lives in (0,1).

it converges to a point, $0$, which is not in your set (0,1).

and so we say that your set is not sequentially compact.

(and so it is not compact.)

edited Apr 25 '16 at 07:45

answered Apr 25 '16 at 07:40

User001

1

1

Oh, this is using that fancy theorem about compact iff secuentially compact, thanks a lot! – José Osorio Apr 25 '16 at 07:45
yes, and the closedness is essential, not just merely the boundedness, which you already have. I.e., a compact set is both closed and bounded. please feel free to up-vote my answer, if you are happy :) – User001 Apr 25 '16 at 07:46
1

how are compactness and sequentially compactness related? – miracle173 Apr 25 '16 at 07:52
great question @miracle173 -- in a metric space sequential compactness and compactness are equivalent. This is not true for a general topological space, however. – User001 Apr 25 '16 at 07:57

score 1 · Answer 4 · answered Apr 25 '16 at 09:19

Yet another argument (which may be useful): we are used to slogan that "in $\mathbb{R}^n$ compact means closed and bounded" (from this you already see that $(0,1)$ not being closed cannot be compact) however compactness is quite different in nature from properties of being bounded and closed:
-being bounded is only with respect to some metric: it requires finer structure on our topological space
-closedness is not intrinsic property: the same set may be closed or not, depending on how you embedd it in some ambient space.
This is not the case for compactness!
-compactness does not require metric structure (only topological)
-compactness is intrinsic (independent from the embedding).
And why I'm talking about this? Since $(0,1)$ is topologically the same as $\mathbb{R}$ so if $(0,1)$ would be compact, so would be $\mathbb{R}$ and for the latter probably you won't have doubts that it is not.

Why can't a open interval in $\mathbb{R}$ be compact?

4 Answers4