# compact sets are closed and bounded

• May 22nd 2011, 09:36 PM
abhishekkgp
compact sets are closed and bounded
Question:
Let $A \subset \mathbb{R}$, then prove that:
$A$ is compact $\Rightarrow A$ is closed and bounded.

approach:
$A$ is compact then $A$ is bounded:
The open intervals $(-n,n), n \in \mathbb{Z}^+$ have union $\mathbb{R}$, so they definitely cover $A$. since $A$ is compact and the above covering is open so it admits a finite subcovering which means $A \subset (-N,N)$ for some $N \in \mathbb{R}$ and hence $A$ is bounded.

now to show that A is closed too.
here is the part where i am stuck:

Let $(x_n)$ be a sequence in $A$ such that $x_n \rightarrow x$. I must show that $x \in A$. i tried to use contradiction method, that is, i assumed that there exists a sequence $(x_n)$ in $A$ with $x_n \rightarrow x$ such that $x \notin A$. now i must exhibit an open covering of $A$ which does NOT admit a finite subcovering in order to bring up a contradiction but i am not able to go further.
• May 22nd 2011, 09:47 PM
Drexel28
Quote:

Originally Posted by abhishekkgp
Question:
Let $A \subset \mathbb{R}$, then prove that:
$A$ is compact $\Rightarrow A$ is closed and bounded.

approach:
$A$ is compact then $A$ is bounded:
The open intervals $(-n,n), n \in \mathbb{Z}^+$ have union $\mathbb{R}$, so they definitely cover $A$. since $A$ is compact and the above covering is open so it admits a finite subcovering which means $A \subset (-N,N)$ for some $N \in \mathbb{R}$ and hence $A$ is bounded.

This is good.

Quote:

now to show that A is closed too.
here is the part where i am stuck:

Let $(x_n)$ be a sequence in $A$ such that $x_n \rightarrow x$. I must show that $x \in A$. i tried to use contradiction method, that is, i assumed that there exists a sequence $(x_n)$ in $A$ with $x_n \rightarrow x$ such that $x \notin A$. now i must exhibit an open covering of $A$ which does NOT admit a finite subcovering in order to bring up a contradiction but i am not able to go further.
There is an easier way to go about this. Let $x\in \mathbb{R}-A$. Then, for each $a\in A$ you can form disjoint open balls $B_a,C_a$ which contain $a,x$ respectively. Note then that evidently $\left\{B_a\right\}_{a\in A}$ is an open cover for $A$ and so by assumption has some finite subcover $B_{a_1},\cdots,B_{a_n}$. Now tell me, why is $C_{a_1}\cap \cdots\cap C_{a_n}$ a neighborhood of $x$ disjoint from $A$?
• May 22nd 2011, 10:08 PM
abhishekkgp
Quote:

Originally Posted by Drexel28
This is good.

There is an easier way to go about this. Let $x\in \mathbb{R}-A$. Then, for each $a\in A$ you can form disjoint open balls $B_a,C_a$ which contain $a,x$ respectively. Note then that evidently $\left\{B_a\right\}_{a\in A}$ is an open cover for $A$ and so by assumption has some finite subcover $B_{a_1},\cdots,B_{a_n}$. Now tell me, why is $C_{a_1}\cap \cdots\cap C_{a_n}$ a neighborhood of $x$ disjoint from $A$?

if $a \in B_{a_1} \cup \ldots \cup B_{a_n}$ then $a \in B_{a_j}$ for some $j$. hence $a \notin C_{a_j}$ and hence $a \notin C_{a_1} \cap \ldots \cap C_{a_n}$.
also, intersection of neighborhoods is a neighborhood. so you have effectively proved that $x \notin A \Rightarrow x \notin \bar{A}$ and it does the job.

but i am still having a doubt. if there exist a sequence $(x_n)$ in $A$ such that $x_n \rightarrow x \notin A$ then $C_{a_1} \cap \ldots \cap C_{a_n}$ will be a singleton namely $\{ x \}$. i am not able to see how the proof rules this out. can you please explain??
• May 22nd 2011, 10:44 PM
abhishekkgp
Using the notations of your post, what i want to say is that what if $B_{a_j}$ is the open interval $(w,x)$ for some $w. if this happens then again $C_{a_1} \cap \ldots \cap C_{a_n}= \{x \}$ whcih is NOT a neighborhood. :( (Headbang)
this is a special example, of course many other such examples are possible.
i don't see how the proof tackles this...
• May 22nd 2011, 11:08 PM
Drexel28
Quote:

Originally Posted by abhishekkgp
Using the notations of your post, what i want to say is that what if $B_{a_j}$ is the open interval $(w,x)$ for some $w. if this happens then again $C_{a_1} \cap \ldots \cap C_{a_n}= \{x \}$ whcih is NOT a neighborhood. :( (Headbang)
this is a special example, of course many other such examples are possible.
i don't see how the proof tackles this...

Well, you should reevaluate what you just said since the finite intersections of open sets is open (this is what makes the set of possible unions of open intervals a topology) and so you can't have that $C_{a_1}\cap\cdots\cap C{a_n}=\{x\}$ since $\{x\}$ isn't open in this topology.
• May 22nd 2011, 11:39 PM
abhishekkgp
Quote:

Originally Posted by Drexel28
Well, you should reevaluate what you just said since the finite intersections of open sets is open (this is what makes the set of possible unions of open intervals a topology) and so you can't have that $C_{a_1}\cap\cdots\cap C{a_n}=\{x\}$ since $\{x\}$ isn't open in this topology.

i agree that finite intersections of neighborhoods of x is again a neighborhood of x.
but if some $B_{a_j}=(w,x)$ then $C_{a_i}=\{x \}$ for some $i$.
if the above seems to be wrong then consider the following:
let $x \in \mathbb{R}-A$.if there exist a sequence $(x_n)$ in $A$ such that $x_n \rightarrow x$. Then any neighborhood of $x$ will intersect $A$. isn't it? how does the proof work here? it means then we have to rule out the possibility of the existence of such a sequence and that is where i am stuck.
• May 23rd 2011, 12:12 AM
Drexel28
Quote:

Originally Posted by abhishekkgp
i agree that finite intersections of neighborhoods of x is again a neighborhood of x.
but if some $B_{a_j}=(w,x)$ then $C_{a_i}=\{x \}$ for some $i$.
if the above seems to be wrong then consider the following:
let $x \in \mathbb{R}-A$.if there exist a sequence $(x_n)$ in $A$ such that $x_n \rightarrow x$. Then any neighborhood of $x$ will intersect $A$. isn't it? how does the proof work here? it means then we have to rule out the possibility of the existence of such a sequence and that is where i am stuck.

I chose the neighborhoods to be disjoint. If you had $B_{a_j}=(w,x)$ then what neighborhood $C_{a_i}$ could be disjoint from it? In particular I was thinking (although a more general idea [Hausdorffness] whose language fit my response better applies) to choose $B_{a_i}=B_{\delta}(a_i)$ and $C_{a_i}=B_{\delta}(x)$ where $\delta=\frac{1}{2}d(x,a_i)$.
• May 23rd 2011, 12:32 AM
abhishekkgp
Quote:

Originally Posted by Drexel28
I chose the neighborhoods to be disjoint. If you had $B_{a_j}=(w,x)$ then what neighborhood $C_{a_i}$ could be disjoint from it? In particular I was thinking (although a more general idea [Hausdorffness] whose language fit my response better applies) to choose $B_{a_i}=B_{\delta}(a_i)$ and $C_{a_i}=B_{\delta}(x)$ where $\delta=\frac{1}{2}d(x,a_i)$.

okay... forget about the $(w,x)$ thing ... my mistake.

still the sequence thingy is bugging me. i will write it again.
let $x \in \mathbb{R} -A$. Assume there exists a sequence $(x_n)$ in A such that $x_n \rightarrow x$. Then $(x_n)$ has a monotonic subsequence $(x_{n_k})$. now if we take the C's as you have mentioned in the above post then $\bigcap C_{x_{n_k}}=\{ x \}$. isn't it? here we don't have a finite number of C's. so the choosing has to be done in a different way. our aim is to show that no such sequence $(x_n)$ exists.can you please clarify this part.
• May 23rd 2011, 02:34 AM
abhishekkgp
Quote:

Originally Posted by Drexel28
I chose the neighborhoods to be disjoint. If you had $B_{a_j}=(w,x)$ then what neighborhood $C_{a_i}$ could be disjoint from it? In particular I was thinking (although a more general idea [Hausdorffness] whose language fit my response better applies) to choose $B_{a_i}=B_{\delta}(a_i)$ and $C_{a_i}=B_{\delta}(x)$ where $\delta=\frac{1}{2}d(x,a_i)$.

after much thought i am starting to see that i was committing a fundamental mistake. sorry for being a blockhead. thanks for being so patient and helping.