# Compactness

• Mar 10th 2010, 09:51 AM
Showcase_22
Compactness
I fell like this question is simple, but I just can't get my head around it:

Quote:

Give an example of a space with two points in which not all compact sets are closed.
Thanks to anyone who replies!
• Mar 10th 2010, 10:43 AM
Focus
Quote:

Originally Posted by Showcase_22
I fell like this question is simple, but I just can't get my head around it:

Thanks to anyone who replies!

Well there are only three topologies you can put on it. The discrete one won't work as all sets are closed, the trivial topology won't work either. The only option is $\{\emptyset,\{a\},\{a,b\}\}$. There is only one set that is open but not closed, and only one sequence in it.
• Mar 10th 2010, 11:06 AM
Showcase_22
What do you mean by "and only one sequence in it"?

Also, is the "open but not closed" set the empty set? I was told that the empty set was simultaneously closed and open, so what's the difference between "simultaneously closed and open" and "open but not closed"?
• Mar 10th 2010, 11:33 AM
Defunkt
If a set is both open and closed, then obviously it can't be open and not closed..
Do you understand why in any topological space $X$, both $X$ and $\phi$ are open and closed?

The set he is talking about is $\{a\}$. Do you see why it is open and not closed? And since it only has one point in it, do you see why it has only one sequence?

Now, can you finish?
• Mar 10th 2010, 12:08 PM
Showcase_22
Quote:

Do you understand why in any topological space , both and are open and closed?
I think I do:

$\phi$ is open since the interior of $\phi$ is empty. If a set is equal to it's interior then it's open.

$\phi$ is closed since all it's boundary points are in $\phi$ since it doesn't have any.

Once we have this, $X^c=\phi$ and since $\phi$ is either open or closed, we get that $X^c$ is also either open or closed.

Quote:

The set he is talking about is . Do you see why it is open and not closed? And since it only has one point in it, do you see why it has only one sequence?
$\{a\}$ is open because it is equal to it's interior. However, $\{a\}$ is surely closed? $\{a\}^c=\{b\}$ (since it's a two point space) and, by the same argument for why $\{a\}$ was open, $\{b\}$ is open. We have a complement of a set as open, so $\{a\}$ must be closed.

Why isn't that right?

Quote:

And since it only has one point in it, do you see why it has only one sequence?
I see why, if you've only got one point any sequence would just be that point repeatedly!

$\{a\}$ has got to be closed since the only sequence is just $a$ repeated, so it has a convergent subsequence. This is just another sequence of just $a$'s, so it converges to $a$.

We know that every convergent sequence has a convergent subsequence iff the set is compact. Therefore $\{a\}$ is compact.
• Mar 10th 2010, 12:16 PM
Focus
Quote:

Originally Posted by Showcase_22
$\{a\}$ is open because it is equal to it's interior. However, $\{a\}$ is surely closed? $\{a\}^c=\{b\}$ (since it's a two point space) and, by the same argument for why $\{a\}$ was open, $\{b\}$ is open. We have a complement of a set as open, so $\{a\}$ must be closed.

Why isn't that right?

Your notion of interior is flawed. Interior is the largest open subset of the set and I (or you) define the open sets with the topology. So when I have {a} in the topology but not {b}, that means that the largest open subset of {b} is the empty set.

A better way to think about open and closed sets is through topology. The whole space and the empty set are open by definition. A set is closed, if and only if its complement is open (in this case $\{b\}=\{a\}^c$ which is not open so {a} is not closed).
• Mar 10th 2010, 12:21 PM
Drexel28
Quote:

Originally Posted by Showcase_22
We know that every convergent sequence has a convergent subsequence iff the set is compact. Therefore $\{a\}$ is compact.

Even more egregious is the above statement. Do not think about compactness as the above unless you are in metric spaces.

As Focus said there are only three topologies on $\{a,b\}$ up to homeomorphism. They are namely $\left\{\varnothing,\{a,b\}\right\},\left\{\varnoth ing,\{a\},\{b\},\{a,b\}\right\}$ and $\left\{\varnothing,\{a\},\{a,b\}\right\}$.

The first is out since every non-open set isn't closed.. The last is out since it is metrizable under the discrete metric (and thus Hausdorff and so every compact subspace is closed). So it remains that we must choose $\left\{\varnothing,\{a\},\{a,b\}\right\}$. Now, clearly $\{a\}$ as a subspace has a finite topology and thus by necessity is compact. But, it is open since it is in the topology but it is not closed since $X-\{a\}=\{b\}$ which is not in the topology.
• Mar 10th 2010, 12:35 PM
Showcase_22
Oh, I think I get it!

So when you write $

\{\emptyset,\{a\},\{a,b\}\}
$
, and define this to be your topology, all the sets in this set are defined to be open. I haven't encountered it written like this before.

So, returning to the question, you could have the set $\{2,5\}$ where $

\{\emptyset,\{2\},\{2,5\}\}
$
is your topology (ie. all the sets defined to be open).

$\{2\}$ is a compact set because every sequence would be of the form $a_n=2$ $\forall n \in \mathbb{N}$. This gives that every subsequence converges since every subsequence $a_{n_i}=2$ (ie. converges to 2).

$\{2\}$ is not closed since $\{2\}^c=\{5\}$ and this is not open since $\{5\}$ is not in the topology (is this the correct way of saying this?).

Alternatively, looking at the definition, a set is closed $\Leftrightarrow$ it's complement is open. Since the complement is not open then our original set is not closed (i'm trying to think of it multiple ways).
• Mar 10th 2010, 12:47 PM
Defunkt
Quote:

Originally Posted by Showcase_22
Oh, I think I get it!

So when you write $

\{\emptyset,\{a\},\{a,b\}\}
$
, and define this to be your topology, all the sets in this set are defined to be open. I haven't encountered it written like this before.

So, returning to the question, you could have the set $\{2,5\}$ where $

\{\emptyset,\{2\},\{2,5\}\}
$
is your topology (ie. all the sets defined to be open).

Everything is fine up to here!

Quote:

$\{2\}$ is a compact set because every sequence would be of the form $a_n=2$ $\forall n \in \mathbb{N}$. This gives that every subsequence converges since every subsequence $a_{n_i}=2$ (ie. converges to 2).
Wrong reasoning, but the correct result. In a general topological space, compactness (Every open cover has a finite subcover) does NOT necessarily ensure sequential compactness (every sequence has a convergent subsequence), for example $X = \{0,1\}^{[0,1]}$ with the product topology.

The right reasoning here is that every open covering of $\{a\}$ is finite, and thus you always have a finite subcover.

Quote:

$\{2\}$ is not closed since $\{2\}^c=\{5\}$ and this is not open since $\{5\}$ is not in the topology (is this the correct way of saying this?).

Alternatively, looking at the definition, a set is closed $\Leftrightarrow$ it's complement is open. Since the complement is not open then our original set is not closed (i'm trying to think of it multiple ways).
Correct.

As a final note, you will probably later learn that in any metric space $(X, d)$, the following 3 conditions are equivalent:
1) X is compact
2) X is sequentially compact
3) X is Limit-point compact (every infinite subset of X has a limit point)

So you would actually be better off not thinking of a general topological space as a generic metric space!
• Mar 10th 2010, 09:46 PM
Showcase_22
Quote:

In a general topological space, compactness (Every open cover has a finite subcover) does NOT necessarily ensure sequential compactness (every sequence has a convergent subsequence), for example with the product topology.
I'm afraid I don't follow this =S

Here compactness does not imply sequential compactness, but I know that a set is compact iff a set is sequentially compact.

Are you saying that, in general, compactness does not imply sequential compactness but a set is compact iff a set is sequentially compact holds in a euclidean space?

Quote:

$

X = \{0,1\}^{[0,1]}
$
with product topology.
Also, what does this mean? I haven't seen this before =S

Is this the cartesian product of a set $\{0,1\}$ with $[0,1]$?

Elements would be like $(0,1), \ \left( 1, \frac{1}{2} \right)$ as in the x value is either 0 or 1 and the y value is any number in $[0,1]$.

If this is right, what is product topology?
• Mar 10th 2010, 10:37 PM
Defunkt
Quote:

Originally Posted by Showcase_22
I'm afraid I don't follow this =S

Here compactness does not imply sequential compactness, but I know that a set is compact iff a set is sequentially compact.

Are you saying that, in general, compactness does not imply sequential compactness but a set is compact iff a set is sequentially compact holds in a euclidean space?

Yes, that is correct, but also for any metric space and not just a Euclidean one (for example $C[[0,1], \mathbb{R}]$, the set of all continuous functions from $[0,1]$ to $\mathbb{R}$).
Though that brings up the question, have you studied topology yet?

Quote:

Also, what does this mean? I haven't seen this before =S

Is this the cartesian product of a set $\{0,1\}$ with $[0,1]$?

Elements would be like $(0,1), \ \left( 1, \frac{1}{2} \right)$ as in the x value is either 0 or 1 and the y value is any number in $[0,1]$.

If this is right, what is product topology?
Well, in set theory, if you have $A,B$ be sets then the notation $B^A$ stands for the set of all functions from A to B.

So generally speaking, $\{0,1\}^{[0,1]}$ is the set of all functions from the unit interval to $\{0,1\}$. Equipping it with the product topology is another way of looking at it from a topological point of view - as a space of uncountably many cartesian products of the set $\{0,1\}$, but now I guess that example is too advanced for what you have learned.

An element $a$ in that set would be of the form $a = \{a_i : a_i \in \{0,1\} \ \forall i \in [0,1]\}$

Also, wikipedia provides a good first read on the product topology, if you're interested. By the way, what course was this question a part of?
• Mar 11th 2010, 12:17 PM
Showcase_22
I'm studying topology as part of my "metric spaces" course. I think it's interesting, i'm just not any good at it =S

This question was from a book I'm reading to try and understand it better. It is helping, the big problem is that it doesn't have the answers in the back of the book!

Anyway, I understand it a lot better since creating this thread. Thanks so much!