Homework 4

This is a place to discuss strategies on HW 4.

1.) Prove that the only subsets of $\mathbb{R}$ that are both open and closed are the whole set $\mathbb{R}$ itself and the empty set $\emptyset$.

My thought is to let $S \subset \mathbb{R}$ that is both closed and open and find a contradiction.

I posted a (somewhat) helpful item in "A Place to share things" or whatever the topic is titled.
I like the idea of proof by contradiction.

I did this proof by supposing $S$ is a nonempty clopen subset of $\mathbb{R}$, and showing that it must equal $\mathbb{R}$. Proof by contradiction is also possible.

So, did you prove that R, is open AND closed first? Then go about showing that there is no other set that can do this?

I was under the impression that we only needed to prove the uniqueness part. I don't think showing $\mathbb{R}$ and $\varnothing$ are clopen would be too difficult, though.

Yes that is how I did it because to prove they are the only sets, I think it is necessary to prove they actually meet the criteria. It is relatively simple though.

So do we need to prove that they are the ONLY subsets of $\mathbb{R}$ that are both open and closed or the fact that they are both open and closed?? I spent the last couple of days thinking it was the latter and now I am questioning my entire proof. Has anyone asked Prof. McClure?

Did you prove that they meet the criteria before proving that they are the only sets? @jmincey

I proved that both $\mathbb{R}$ and $\emptyset$ are open and closed and then proved that they are the only ones that are both open and closed. Mark never said if we could assume that we could assume $\mathbb{R}$ and $\emptyset$ are open and closed so I proved it.

Yeah, I proved it too but I have not yet proved that they are the only ones...Did contradiction work for you? I am not sure how to proceed. @jmincey

@ediazloa seems to me that $\mathbb{R}$ and $\emptyset$ being clopen is part of the hypothesis (though I could be wrong) Contradiction worked for me, though it took some massaging. After I let $S\subset\mathbb{R}$ be clopen my next move was to let $a\in S$ and $b\in \{\mathbb{R} - S\}$ and defined an interval $A = \{x \in [a,b] : [a, x] \subseteq S\}$. It is also important to realize that $S, \{\mathbb{R} - S\}$ are both open sets...

For proving that $\mathbb{R}$ and $\emptyset$ are both closed and open do you think that it is sufficient to prove that one of them is both open and closed and then used the fact that they are each other's compliment. With the properties that if A is closed then $A^C$ is open and viceversa,

Also I asked Mark whether we had to prove that R and emptyset was both open and closed and he said that we indeed had to.

I have also been working with the idea that $\mathbb{R}$and the empty sets are compliments and that proving one is both closed and empty also proves the other. I think the rest where we prove they are the only ones that are both may have o do with compliments also.

I think so too. As people have mentioned we could use contradiction, then letting S be some proper subset that is both open and closed. By the previous logic it's compliment must then also be open and all we would need to do is to prove that either one interior or limit point exists in both for a contradiction, since they wouldn't be compliments anymore.

The only problem with that is that there are infinitely many possible sets and we have to show that only the two mentioned are both open and closed, so how do we prove an infinite number of sets have to be open or closed?

We don't. We are trying to prove that there doesn't exist any other set than $\mathbb{R}$ and $\emptyset$. Most people seem to like to use contradiction and to do that one could start by assuming that there exist a proper subset $a\subset \mathbb{R}$ that is both open and closed and then find some contradiction that shows that this set doesn't exist.

Did you use idea of supremum or infemum? That may be applicable next.. @jmacdona

This right here is an underrated post.

@jmacdona It might just be your notation but I don't understand what you are trying to define to be your subset $A$. You have $a\in S$ and $b\in S^c$ but then you define $A=\{ x\in [a,b] \mid [a,x] \subseteq S \}$. This notation seems to only make sense if $b>a$ or else the intervals wouldn't make sense. If this was the case, this set is really just $A = \{ x\in S \mid b>x\ge a\}$. Is that what you were going for?