I need to construct a set (Topology question)

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  • #1

Homework Statement

Construct a set S in R such that
- S
- interior of S
- Closure of S
- Closure of the interior of S
- Interior of the closure of S

Are all distinct from each other.

Homework Equations

I don't think there should be any.

The Attempt at a Solution

I 'm pretty sure that the set has something to do with the rational numbers Q, but I'm stuck at "closure of the interior of S" and "interior of the closure of S".

So far, my attempt has been:

- S: The rational numbers Q
- int S: Empty set
- cl S: R
- cl (int S) : ?
- int (cl S) : ?

Just wondering, does this work? If not, is there a better set for this type of question?

Answers and Replies

  • #2
Using the rationals is clever, but your example in its original form doesn't quite work. Instead of trying to use all of [tex] \mathbb{Q} [/tex], you should restrict attention to just a convenient portion of it, like, for example, [tex] \mathbb{Q} [/tex] restricted to some compact set. (\hint)

Restricting [tex] \mathbb{Q} [/tex] in this way almost gives you the set you need, but to finish the construction, you need to add on another "piece." (This last part isn't hard; almost any additional set will do, I think.) Sorry to be so vague, but I don't want to give too much away.
  • #3
Thank you for your hint, I just wonder if it's possible to continue to work on the solution along the way while getting feedbacks from forum members...

So I just need to restrict S to some smaller/specific numbers? Something like 0 to 1?

And about the additional requirements for the set, does it involve some sort of union/intersection?
  • #4
So I just need to restrict S to some smaller/specific numbers? Something like 0 to 1?

Exactly. In fact, the set [tex] \mathbb{Q} \cap [0,1] [/tex] is just what I had in mind.

The only problem with this is that the interior is the same as the closure of the interior (they're both the null set). To fix this, just take the union of this set with some other set whose interior is different from the closure of its interior.
  • #5
Nice! That's one step closer to the answer, but somehow I can't distinguish the difference between int S and cl(int S)...

Say, if I union Q /\ [0, 1] with a set {any number than 0 and 1}, would that work?
So would it be that the interior is the set {..}, and the closure of the interior is R - {..}?

Err... so cl S is still R, but what about int(cl S)? That's int(R), but how should I represent it?
  • #6
Well, the interior of [tex] \mathbb{Q} [/tex] or [tex] \mathbb{Q} \cap [0,1] [/tex] is just [tex] \emptyset [/tex], and the closure of the empty set is the empty set. (If that confuses you, remember that the definition of the closure of [tex] S [/tex] is the set of all points [tex] p [/tex] such that any neighborhood of [tex] p [/tex] has a nontrivial intersection with [tex] S [/tex]. Any set intersected with the null set is trivial. Thus, the closure of the null set is empty.) However, the closure of [tex] \mathbb{Q} \cap [0,1] [/tex] is just [tex] [0,1] [/tex], not all of [tex] \mathbb{R} [/tex]. (Why?)

How about this: Let [tex] I = [a,b] [/tex] be an arbitrary closed interval, and let [tex] S = \mathbb{Q} \cap [0,1] [/tex]. The interior of [tex] I [/tex] is the open interval [tex] (a,b) [/tex], right? What's the closure of [tex] (a,b) [/tex] (i.e., the closure of the interior of [tex] I [/tex])? Using this information, can you see why the set [tex] S \cup I [/tex] will work if [tex] a [/tex] and [tex] b [/tex] are both greater than 1?
  • #7
Right! I get it now.

If I set S to (Q /\ [0,1]) \/ [2,3], then int S = (2,3), cl S = [0,1], int (cl S) = (0,1), and cl (int S) = [2,3]?

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