# Thread: compact sets are closed and bounded

1. ## compact sets are closed and bounded

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

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

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

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

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

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

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

Let $\displaystyle (x_n)$ be a sequence in $\displaystyle A$ such that $\displaystyle x_n \rightarrow x$. I must show that $\displaystyle x \in A$. i tried to use contradiction method, that is, i assumed that there exists a sequence $\displaystyle (x_n)$ in $\displaystyle A$ with $\displaystyle x_n \rightarrow x$ such that $\displaystyle x \notin A$. now i must exhibit an open covering of $\displaystyle A$ 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 $\displaystyle x\in \mathbb{R}-A$. Then, for each $\displaystyle a\in A$ you can form disjoint open balls $\displaystyle B_a,C_a$ which contain $\displaystyle a,x$ respectively. Note then that evidently $\displaystyle \left\{B_a\right\}_{a\in A}$ is an open cover for $\displaystyle A$ and so by assumption has some finite subcover $\displaystyle B_{a_1},\cdots,B_{a_n}$. Now tell me, why is $\displaystyle C_{a_1}\cap \cdots\cap C_{a_n}$ a neighborhood of $\displaystyle x$ disjoint from $\displaystyle A$?

3. Originally Posted by Drexel28
This is good.

There is an easier way to go about this. Let $\displaystyle x\in \mathbb{R}-A$. Then, for each $\displaystyle a\in A$ you can form disjoint open balls $\displaystyle B_a,C_a$ which contain $\displaystyle a,x$ respectively. Note then that evidently $\displaystyle \left\{B_a\right\}_{a\in A}$ is an open cover for $\displaystyle A$ and so by assumption has some finite subcover $\displaystyle B_{a_1},\cdots,B_{a_n}$. Now tell me, why is $\displaystyle C_{a_1}\cap \cdots\cap C_{a_n}$ a neighborhood of $\displaystyle x$ disjoint from $\displaystyle A$?
if $\displaystyle a \in B_{a_1} \cup \ldots \cup B_{a_n}$ then $\displaystyle a \in B_{a_j}$ for some $\displaystyle j$. hence $\displaystyle a \notin C_{a_j}$ and hence $\displaystyle a \notin C_{a_1} \cap \ldots \cap C_{a_n}$.
also, intersection of neighborhoods is a neighborhood. so you have effectively proved that $\displaystyle x \notin A \Rightarrow x \notin \bar{A}$ and it does the job.

but i am still having a doubt. if there exist a sequence $\displaystyle (x_n)$ in $\displaystyle A$ such that $\displaystyle x_n \rightarrow x \notin A$ then $\displaystyle C_{a_1} \cap \ldots \cap C_{a_n}$ will be a singleton namely $\displaystyle \{ x \}$. i am not able to see how the proof rules this out. can you please explain??

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

5. Originally Posted by abhishekkgp
Using the notations of your post, what i want to say is that what if $\displaystyle B_{a_j}$ is the open interval $\displaystyle (w,x)$ for some $\displaystyle w<x$. if this happens then again $\displaystyle C_{a_1} \cap \ldots \cap C_{a_n}= \{x \}$ whcih is NOT a neighborhood.
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 $\displaystyle C_{a_1}\cap\cdots\cap C{a_n}=\{x\}$ since $\displaystyle \{x\}$ isn't open in this topology.

6. 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 $\displaystyle C_{a_1}\cap\cdots\cap C{a_n}=\{x\}$ since $\displaystyle \{x\}$ isn't open in this topology.
i agree that finite intersections of neighborhoods of x is again a neighborhood of x.
but if some $\displaystyle B_{a_j}=(w,x)$ then $\displaystyle C_{a_i}=\{x \}$ for some $\displaystyle i$.
if the above seems to be wrong then consider the following:
let $\displaystyle x \in \mathbb{R}-A$.if there exist a sequence $\displaystyle (x_n)$ in $\displaystyle A$ such that $\displaystyle x_n \rightarrow x$. Then any neighborhood of $\displaystyle x$ will intersect $\displaystyle A$. 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.

7. Originally Posted by abhishekkgp
i agree that finite intersections of neighborhoods of x is again a neighborhood of x.
but if some $\displaystyle B_{a_j}=(w,x)$ then $\displaystyle C_{a_i}=\{x \}$ for some $\displaystyle i$.
if the above seems to be wrong then consider the following:
let $\displaystyle x \in \mathbb{R}-A$.if there exist a sequence $\displaystyle (x_n)$ in $\displaystyle A$ such that $\displaystyle x_n \rightarrow x$. Then any neighborhood of $\displaystyle x$ will intersect $\displaystyle A$. 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 $\displaystyle B_{a_j}=(w,x)$ then what neighborhood $\displaystyle C_{a_i}$ 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 $\displaystyle B_{a_i}=B_{\delta}(a_i)$ and $\displaystyle C_{a_i}=B_{\delta}(x)$ where $\displaystyle \delta=\frac{1}{2}d(x,a_i)$.
I chose the neighborhoods to be disjoint. If you had $\displaystyle B_{a_j}=(w,x)$ then what neighborhood $\displaystyle C_{a_i}$ 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 $\displaystyle B_{a_i}=B_{\delta}(a_i)$ and $\displaystyle C_{a_i}=B_{\delta}(x)$ where $\displaystyle \delta=\frac{1}{2}d(x,a_i)$.
okay... forget about the $\displaystyle (w,x)$ thing ... my mistake.
let $\displaystyle x \in \mathbb{R} -A$. Assume there exists a sequence $\displaystyle (x_n)$ in A such that $\displaystyle x_n \rightarrow x$. Then $\displaystyle (x_n)$ has a monotonic subsequence $\displaystyle (x_{n_k})$. now if we take the C's as you have mentioned in the above post then $\displaystyle \bigcap C_{x_{n_k}}=\{ x \}$. 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 $\displaystyle (x_n)$ exists.can you please clarify this part.
I chose the neighborhoods to be disjoint. If you had $\displaystyle B_{a_j}=(w,x)$ then what neighborhood $\displaystyle C_{a_i}$ 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 $\displaystyle B_{a_i}=B_{\delta}(a_i)$ and $\displaystyle C_{a_i}=B_{\delta}(x)$ where $\displaystyle \delta=\frac{1}{2}d(x,a_i)$.