# Verifying validity of a statement involving infinite compact sets

• Sep 29th 2010, 08:29 AM
Pinkk
Verifying validity of a statement involving infinite compact sets
Theorem: Show that an infinite set $\displaystyle S$ is compact if and only if every infinite subset of $\displaystyle S$ has an accumulation point that lies in $\displaystyle S$.

My attempt of proof:

Suppose $\displaystyle S$ is infinite and compact but that there exists an infinite subset that does not contain any accumulation points in $\displaystyle S$. Call this subset $\displaystyle S'$ and let an accumulation point of this subset be $\displaystyle a$. Since $\displaystyle a$ is an accumulation point of $\displaystyle S'$, there is a sequence of points in $\displaystyle S'$ such that the limit of that sequence is $\displaystyle a$. Clearly this sequence lies in $\displaystyle S$ as well since $\displaystyle S' \subset S$, so we have that $\displaystyle a$ is an accumulation point of $\displaystyle S$, and since $\displaystyle S$ is compact, $\displaystyle a\in S$, which is a contradiction. So all infinite subsets of $\displaystyle S$ have an accumulation point which lies in $\displaystyle S$.

Now suppose every infinite subset of $\displaystyle S$ has an an accumulation point in $\displaystyle S$ and that $\displaystyle S$ is not compact. Then $\displaystyle S$ is either unbounded or not closed. If $\displaystyle S$ is unbounded, then there exists a sequence in $\displaystyle S$, say $\displaystyle \{x_{n}\}$, such that for all $\displaystyle n\in \mathbb{N}$, $\displaystyle x_{n} \ge n$, so $\displaystyle \lim x_{n} = \infty$. So let $\displaystyle S''$ be the set of these sequence elements. By construction, this is an infinite set, yet it has no accumulation point since no matter the ordering of the terms, there is no such point $\displaystyle b$ such that infinitely many sequence terms lie arbitrary close to $\displaystyle b$. This contradictions our hypothesis, so $\displaystyle S$ is bounded. Suppose $\displaystyle S$ is not closed. Then there exists a point $\displaystyle c$ such that $\displaystyle c\in \partial S$ yet $\displaystyle c\not\in S$. Since $\displaystyle c\in \partial S$, there exists a sequence of points in $\displaystyle S$, say $\displaystyle \{y_{n}\}$ such that $\displaystyle \lim y_{n} = c$. Let $\displaystyle S'''$ be the set of these sequence elements. By construction, this is an infinite set, yet its only accumulation point is $\displaystyle c$ since no matter the ordering of the terms, $\displaystyle c$ is the only point where infinitely many terms of the sequence are arbitrarily close to. Again, this contradicts the hypothesis, so $\displaystyle S$ must be compact. $\displaystyle \blacksquare$

Is this proof valid, the only part of the proof I am really unsure of is in the second paragraph with the construction of those sequences. Is my claim valid that no matter the ordering, there are no other accumulation points then the ones I have specified? It seems obvious/true to me (I think) but is it a false claim or is it a claim that needs proof? Thank you.
• Sep 29th 2010, 10:02 AM
Plato
Quote:

Originally Posted by Pinkk
Theorem: Show that an infinite set $\displaystyle S$ is compact if and only if every infinite subset of $\displaystyle S$ has an accumulation point that lies in $\displaystyle S$.

Suppose $\displaystyle S$ is infinite and compact but that there exists an infinite subset that does not contain any accumulation points in $\displaystyle S$. Call this subset $\displaystyle S'$ and let an accumulation point of this subset be $\displaystyle a$.

Look at the statements in blue and red.
You can not have both ways: either the set has an accumulation point or it does not.

So try again. If it has no accumulation point, then the is a open infinite cover which does have a finite sub-cover.
• Sep 29th 2010, 10:05 AM
Pinkk
Well, since $\displaystyle S'$ is infinite and is a subset of a bounded set, the subset itself is bounded, so it MUST have an accumulation point (which is another theorem with another proof). I'm trying to prove that this accumulation point must lie in $\displaystyle S$, so I'm assuming to the contrary that the accumulation point of $\displaystyle S'$, which exists, does not lie in $\displaystyle S$ which will lead to a contradiction that $\displaystyle S$ is compact.
• Sep 29th 2010, 10:15 AM
Plato
Quote:

Originally Posted by Pinkk
Theorem: Show that an infinite set $\displaystyle S$ is compact if and only if every infinite subset of $\displaystyle S$ has an accumulation point that lies in $\displaystyle S$.

Read what you posted very carefully. If you already know from another theorem every infinite bounded set has a limit point, then there is nothing to prove in that direction. So you have wasted a lot of words.
• Sep 29th 2010, 10:19 AM
Pinkk
Yes, but that theorem says nothing about where that accumulation point lies. Yes, every bounded infinite set has an accumulation point, so every infinite subset of a compact set has an accumulation point, but it is not immediate that these accumulation points lie in $\displaystyle S$ for ALL infinite subsets. So I assume to the contrary that there is one infinite subset that has a limit point NOT in $\displaystyle S$ and arrive at a contradiction to the hypothesis that $\displaystyle S$ is bounded AND closed.
• Sep 29th 2010, 10:29 AM
Plato
Quote:

Originally Posted by Pinkk
Yes, but that theorem says nothing about where that accumulation point lies.

Compact sets are closed.
• Sep 29th 2010, 12:14 PM
Pinkk
The other theorem makes no assumption about being closed, only bounded and infinite. I guess that an alternate proof to the current theorem could be that the accumulation point belongs to $\displaystyle cl(S')$ and then prove that $\displaystyle cl(S') \subset S$ by assuming to the contrary and deriving a contradiction (or directly works just as well). In any case, I have to prove that the accumulation point of an infinite subset, making no assumption about whether this subset is closed or not, of $\displaystyle S$ has an accumulation point that lies in $\displaystyle S$.
• Sep 29th 2010, 12:31 PM
Plato
Look you are going in circles.
If a set $\displaystyle S$ is compact then it is closed and bounded.
By the B-W theorem any infinite subset has a limit point in $\displaystyle S$.

The other way.
Suppose that every infinite subset of $\displaystyle S$ has a limit point in $\displaystyle S$.
If $\displaystyle \left\{ {O_n } \right\}$ is an infinite open cover for $\displaystyle S$ which has no finite sub-cover work for a contradiction.