Real Analysis: Open and Closed Sets

• Oct 6th 2008, 06:27 PM
ziggychick
Real Analysis: Open and Closed Sets
I have 2 questions hopefully someone can help with.

1) Is R the only open set containing Q? Prove or give an example of a set containing Q.

I do believe R is the only open set containing Q but I have no idea how to prove this.

2) Is the set of integers a closed set? I'm pretty sure it is, since the boundary points are the integers. But is is also open?

Thanks so much for the help!
• Oct 6th 2008, 07:04 PM
robeuler
Quote:

Originally Posted by ziggychick
I have 2 questions hopefully someone can help with.

1) Is R the only open set containing Q? Prove or give an example of a set containing Q.

I do believe R is the only open set containing Q but I have no idea how to prove this.

2) Is the set of integers a closed set? I'm pretty sure it is, since the boundary points are the integers. But is is also open?

Thanks so much for the help!

First number 2: The integers are not open. Any epsilon neighborhood of an integer will contain points not in the set.

For number 1: I am not entirely sure, but you could try taking M such that Q is contained in M is contained in R. Suppose M is open. Doesn't this imply that M=R? (if M is open then for any point m in M there is an epsilon neighborhood around m contained in M. Use the density of the irrationals to show that M=R)
• Oct 7th 2008, 12:18 AM
Opalg
Quote:

Originally Posted by ziggychick
Is R the only open set containing Q? Prove or give an example of a set containing Q.

What can you say about the set consisting of all the real numbers except $\displaystyle \sqrt2$ ?
• Oct 7th 2008, 08:29 PM
ziggychick
riight!
The set (-infty, rt(2)) union (rt(2), +infty) is open and contains all of Q.
Thanks guys!
• Oct 9th 2008, 11:09 AM
bkarpuz
Quote:

Originally Posted by ziggychick

2) Is the set of integers a closed set? I'm pretty sure it is, since the boundary points are the integers. But is is also open?

As far as I remember, in any topology, sets including a single point are closed.
Therefore, union of these sets are also closed. (Thinking)
• Oct 9th 2008, 11:33 AM
Moo
Quote:

Originally Posted by bkarpuz
As far as I remember, in any topology, sets including a single point are closed.

Why ? (Worried)
Quote:

Therefore, union of these sets are also closed. (Thinking)
So it should be a finite union of these sets, isn't it ? (if we consider the complement of the union..)
• Oct 9th 2008, 11:48 AM
bkarpuz
Quote:

Originally Posted by Moo
So it should be a finite union of these sets, isn't it ?

yap must be finite otherwise $\displaystyle \mathrm{Q}$is a contrary example.

Also for the first one, I might be wrong too. (Crying)
• Oct 9th 2008, 12:36 PM
Jhevon
see here
• Oct 9th 2008, 12:41 PM
Moo
Quote:

Originally Posted by Jhevon
see here

In the discrete topology, of course it's true ! :)
Since $\displaystyle \tau=\{ \{a\} ~:~ a \in X \}$

But if you consider the topology over $\displaystyle \mathbb{R}$ : $\displaystyle \tau=\{ \emptyset~,~\mathbb{R}~,~ [0,1] \}$, it's clear that the singletons of $\displaystyle \mathbb{R}$ are nor open, nor closed sets (Surprised) (it's the example I gave to bkarpuz >.<)
So it cannot be true for any topology
• Oct 9th 2008, 12:43 PM
Jhevon
Quote:

Originally Posted by Moo
In the discrete topology, of course it's true ! :)
Since $\displaystyle \tau=\{ \{a\} ~:~ a \in X \}$

But if you consider the topology over $\displaystyle \mathbb{R}=\{ \emptyset~,~\mathbb{R}~,~ [0,1] \}$, it's clear that the singletons of $\displaystyle \mathbb{R}$ are nor open, nor closed sets (Surprised) (it's the example I gave to bkarpuz >.<)

yes, i was responding to what bkarpuz said about "any" topology. his claim was not true for discrete topologies.

anyway, i know nothing about topology, so i shouldn't even be responding to anything here (Worried)
• Oct 9th 2008, 12:46 PM
Moo
Quote:

Originally Posted by Jhevon
yes, i was responding to what bkarpuz said about "any" topology. his claim was not true for discrete topologies.

anyway, i know nothing about topology, so i shouldn't even be responding to anything here (Worried)

CaptainBlack would have warned you to quote the post you're replying to ! (Rofl)

Anyway, it's true for discrete topologies, not "not true" :p
And why do you say you know nothing ? o.O "open" is a topological concept ! :D
Learn a little about it, it's interesting ! :)

Note (mainly to bkarpuz) : in the usual topology over R, a singleton is an open set.
• Oct 9th 2008, 12:55 PM
Jhevon
Quote:

Originally Posted by Moo
CaptainBlack would have warned you to quote the post you're replying to ! (Rofl)

haha, yeah

Quote:

Anyway, it's true for discrete topologies, not "not true" :p
And why do you say you know nothing ? o.O "open" is a topological concept ! :D
i still say i know nothing about topology. we only did a few stuff on it, and i was just about to finish wrapping my head around the concepts in chapter 1 when we stopped.

Quote:

Learn a little about it, it's interesting ! :)
too interesting for me at the moment :D
• Oct 9th 2008, 01:20 PM
Laurent
Quote:

Originally Posted by bkarpuz
As far as I remember, in any topology, sets including a single point are closed.

This is true in Hausdorff spaces - or "separated spaces" -, i.e. if you assume that two distinct points always possess disjoint neighbourhoods.

This is not necessary and sufficient, but nevertheless fairly general.
In fact, the Wikipedia says that the original definition of a topology by Hausdorff included this separation condition. It also gives a pun to remember the definition of Hausdorff space that I let you read by yourselves... (first paragraph)
• Oct 9th 2008, 09:32 PM
bkarpuz
Quote:

Originally Posted by Jhevon
yes, i was responding to what bkarpuz said about "any" topology. his claim was not true for discrete topologies.

anyway, i know nothing about topology, so i shouldn't even be responding to anything here (Worried)

I dont think so, it is helpful to point out a mistake. (Wink)