True/false questions about open/closed sets

[utleysthrow]

Homework Statement

although I'm supposed to provide explanations for these, i just want to see if my intuition about them is correct..

1) For any set A \subseteq R, \overline{A}^{c} is open

2) If a set A has an isolated point, it cannot be an open set

3) Set A is closed if and only if \overline{A} = A

4) If A is a bounded set, s=sup A is a limit point of A

5) Every finite set is closed

6) An open set that contains every rational #s must necessarily be all of R.

Homework Equations

The Attempt at a Solution

1) True
2) True
3) True
4) True
5) False
6) False

 

[rasmhop]

I'm assuming we're only working with subsets of the real line.

I agree with your answers to 1,2,3,6. For 5 I disagree as all one-point sets are closed and finite unions of closed sets are closed. As for 4 I agree for some definitions of limit points (I have seen different definitions in different books). If you require that all neighborhoods of x intersect A in some place OTHER than x, then I disagree since {1} is bounded, sup{1} =1, but 1 is not a limit point of {1}. If you only require that every neighborhood of x intersect A, then I agree.

 

[utleysthrow]

Yeah, you're right for 4 as well, since I'm using the first definition you mentioned. The set doesn't have to be "continuous".

 

[utleysthrow]

Hmmm... I'm thinking that maybe 6 is true.

6) An open set that contains every rational #s must necessarily be all of R.

All rational #s by themselves are closed and so the (countably) finite union of them would be closed... but is it possible to construct an "open" set that just contains rational #s?

Edit: actually... regardless of whether it's possible to create an open set containing rational numbers, any set containing rational numbers should contain irrationals... right? Sorry I'm not sure how to explain it better.

 

[lanedance]

i assume we're talking about open as a subset of R here?

Hmmm... I'm thinking that maybe 6 is true.

All rational #s by themselves are closed and so the (countably) finite union of them would be closed... but is it possible to construct an "open" set that just contains rational #s?

I'm thinking its true as well, as any real number can be approximated abritrarily close by a rational, ie. the rationals are dense in the reals. So any open interval of \mathbb{R} must contain rationals and irrational

Edit: actually... regardless of whether it's possible to create an open set containing rational numbers, any set containing rational numbers should contain irrationals... right? Sorry I'm not sure how to explain it better.

what about {1/3} no irrationals there...

 

[lanedance]

the set of all rational numbers \mathbb{Q} is not open in \mathbb{R} either, as any neighbourhood will contain an irrational which is not in \mathbb{Q}

 

[utleysthrow]

Oops, I meant to say a set containing "every" rational numbers, as the question asks.

(And yes I'm talking about subsets of R.)

As you said, the rationals are dense in the reals, so R=\overline{Q}. So an open set containing every rationals also contains every irrationals, since any open set contains rationals and irrationals (and hence the reals)

 

[rasmhop]

(\sqrt{2},\infty) and (-\infty,\sqrt{2}) are open so
U=(-\infty,\sqrt{2})\cup(\sqrt{2},\infty) = \mathbb{R} - \{\sqrt{2}\}
is open and contains all rationals, but U is not \mathbb{R}.

 

[lanedance]

good point, missed that