# Thread: Closed, bounded and perfect sets

1. ## Closed, bounded and perfect sets

Hello everyone...

I've recently came accross an analysis text that includes in one of the examples the following:

S is the set of all complex numbers (that is, R2)
T is the set of all real integers (Z)

the sets are:
CLOSED OPEN PERFECT BOUNDED
S YES YES YES NO
T YES NO NO NO

Question: How is it that 1) S and T are closed, how can they be closed but not bounded? 2) S is closed and open at the same time?

Any help is much appreciated...

2. Originally Posted by rebghb
Hello everyone...

I've recently came accross an analysis text that includes in one of the examples the following:

S is the set of all complex numbers (that is, R2)
T is the set of all real integers (Z)

the sets are:
CLOSED OPEN PERFECT BOUNDED
S YES YES YES NO
T YES NO NO NO

Question: How is it that 1) S and T are closed, how can they be closed but not bounded? 2) S is closed and open at the same time?

Any help is much appreciated...

This is very misleading. First of all openess is relative to the space you're in.

For example, you would agree with me that as a metric space (or a topological space...whichever is more up your ally) $\mathbb{R}$ (with the usual metric/topology) considers $\{3\}$ closed. But, if you consider $\{3\}$ as being a subspace in it's own right, then it is open in itself. You can't be open. You have to be open in a specific space.

I'm being too finicky. You obviously mean these to be subsets of $\mathbb{R}^2$ with the usual topology.

$S$- Every space is open and closed within itself, it's the definition of a topology! Or, if you're just doing metric spaces note that it's closed since any limit point would have to be a point of the space and thus automatically in it. Also, any point is an interior point since any neighborhood is contained within the space! It's perfect for the same reason $D(\mathbb{R}^2)=\left\{x\in\mathbb{R}^2:x\text{ is a limit point of }\mathbb{R}^2}\subseteq\mathbb{R}^2$. But, $D(\mathbb{R}^2)\supseteq\mathbb{R}^2$ too for if $x\in\mathbb{R}^2$ then taking any neighorhood of $x$ will hit another point of $\mathbb{R}^2-\{x\}$. It's obviously not bounded under the usual metric.

$T$. I'll assume you mean $\mathbb{Z}\subseteq\mathbb{R}$. It's closed since given any $x\in\mathbb{R}$ we have that $B_{\frac{1}{4}}(x)\cap\mathbb{Z}$ can contain at most one point, and so $x$ isn't a limit point. It follows that $\mathbb{Z}$ has no limit points and so trivially contains them. $\mathbb{Z}$ is not open for the easy reason that $\mathbb{R}$ is connected and $\varnothing\subset\mathbb{Z}\subset\mathbb{R}$ (among other reasons). It's not perfect since it is non-empty while it's set of limit points is (or because every pefect subset of $\mathbb{R}$ is uncountable). Finally, it is also clearly nt bounded.

3. Well I'm new to all of this branch in math, so to see if I get it right, we can say that that set R (as a subset of R^2) is Open because we're assuming that it is beign studied with respect to R^2... It is not Closed because the limit points may belong to its complement in R^2, it is not perfect ovbiously because it's open(not closed), and it is not bounded...

4. Originally Posted by rebghb
Well I'm new to all of this branch in math, so to see if I get it right, we can say that that set R (as a subset of R^2) is Open because we're assuming that it is beign studied with respect to R^2... It is not Closed because the limit points may belong to its complement in R^2, it is not perfect ovbiously because it's open(not closed), and it is not bounded...
Hmm, it seems you have the first part backwards. Since you are new I'll try to ease into the topic.

When you say $\mathbb{R}$ as a subset of $\mathbb{R}^2$ that makes no sense. But, you can think of $\mathbb{R}$ as being embedded in $\mathbb{R}$ in the sense that for all intents and purposes it is (in this case) identical to $\mathbb{R}\times\{0\}$.

Thus, with that in mind it is closed because it has no limit points. To see this all you need to do is realize that if $x\notin\mathbb{R}\times\{0\}$...well, for the sake of ease let us think instead about $\mathbb{C}$ instead of $\mathbb{R}^2$ for a second (they are equivalent in the sense that I alluded to earlier). Then, $\mathbb{R}\times\{0\}$ is now merely $\left\{z\in\mathbb{C}:\text{Im}(z)=0\right\}$. So, to see that this set is closed we can show it's complement is open. So, let $w\notin\left\{z\in\mathbb{C}:\text{Im}(z)=0\right\ }$. Then, $w=a+bi,\text{ }b\ne 0$. So then, if you give me any other $w'=a'+b'i\in\mathbb{C}$ I know the distance between the two is $\sqrt{(a-a')^2+(b-b')^2}$. But, clearly this distance is greater than $|b-b'|$, right?

So, how does that help us? Well, since we assumed that $w\notin\left\{z\in\mathbb{C}:\text{Im}(z)=0\right\ }$ we said that $w=a+bi,\text{ }b>0$. So, if you give me ANY $w'=a'+b'i\in\left\{z\in\mathbb{C}:\text{Im}(z)=0\r ight\}$ the distance between the two (as said earlier) is greater than $|b-b'|$. But, $b'=0$! So, the distance between any element of $\left\{z\in\mathbb{C}:\text{Im}(z)=0\right\}$ and $w$ is greater than $|b|>0$!

So, if we look at all the things $\frac{|b|}{2}=\delta$ away from $w$ (this is denoted $B_{\delta}(w)$) this is an open set, and no element of $\left\{z\in\mathbb{C}:\text{Im}(z)=0\right\}$ can be in it! So, $B_{\delta}(w)$ is contained entirely within the complement of that set. Thus, since $w$ was arbitrary this means that, by definition of openess in metric spaces, the complement of our set is open. So! Our set must be closed.

The connectedness statement in the last post was inappropriate. To see that our set isn't open all you have to realize is that given some $a+bi\in\left\{z\in\mathbb{C}:\text{Im}(z)=0\right\ }$ and any $\delta>0$ choosing $a+\frac{\delta}{2}i$ we see that $\left|a+bi-a+\frac{\delta}{2}i\right|=\left|bi-\frac{\delta}{2}i\right|=\frac{\delta}{2}$ (since a+bi was in our set we had to have that b=0). Thus, this $a+\frac{\delta}{2}i$ is in $B_{\delta}(a+bi)$ and in the complement of our set.So no matter how small of a region you look around any point in our set there will be points not in our set. So, it can't be open.

It's isn't perfect since (using a modified argument as in the last paragraph) it has no limit points. (this also proves that it's closed). And yes, you are right that it isn't bounded.

5. Thanks for this amazing reply, point very well made! And thanks for confirming that I've got at least one of the 4 correct

By the way, I forgot to mention that in the text I read IT DID clarify, before classifying the two sets, that the latters were in R^2... Having not noticed that I asked for help, but your post cleared out a WHOOLE lot of information.

Thanks again.

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