Thread: 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!

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)

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 ?

The set (-infty, rt(2)) union (rt(2), +infty) is open and contains all of Q.
Thanks guys!

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.

Originally Posted by bkarpuz

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

Why ?

Therefore, union of these sets are also closed.

So it should be a finite union of these sets, isn't it ? (if we consider the complement of the union..)

Originally Posted by Moo

So it should be a finite union of these sets, isn't it ?

yap must be finite otherwise is a contrary example.

Also for the first one, I might be wrong too.

see here

Originally Posted by Jhevon

see here

In the discrete topology, of course it's true !
Since

But if you consider the topology over : , it's clear that the singletons of are nor open, nor closed sets (it's the example I gave to bkarpuz >.<)
So it cannot be true for any topology

Originally Posted by Moo

In the discrete topology, of course it's true !
Since

But if you consider the topology over , it's clear that the singletons of are nor open, nor closed sets (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

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

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

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



Note (mainly to bkarpuz) : in the usual topology over R, a singleton is an open set.

Originally Posted by Moo

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

haha, yeah

Anyway, it's true for discrete topologies, not "not true"
And why do you say you know nothing ? o.O "open" is a topological concept !

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.

Learn a little about it, it's interesting !

too interesting for me at the moment

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)

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

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