Every infinite set has a countable dense subset?

In summary: Now suppose x is not in A. There is a point in X that's a limit point of A and not in A. That's impossible, right? So x must be in A. That's what we want to show.
  • #1
oma12
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Homework Statement



Prove that every infinite set has a countable dense subset.

Homework Equations





The Attempt at a Solution



I have almost no idea how to solve this problem of my analysis homework. I was thinking that i need to show that there is a countable subset that has all points of the initial set as limit points. So i was thinking of saying that that any open infinite set can be written as a countable union of subsets or like neighbourhoods, and using something like a pigeon hold principle to say that each of these neighbourhoods must contain some points in the countable subset. but we have countable nbhds and countable points, i don't think the pigeonhole principle works here... I'm probably way off the solution... ANY HELP?
 
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  • #2
Not every metric space has a countable dense subset. Google the "long line". Only separable spaces have one. If it's a subset of the reals then you are ok.
 
  • #3
ok, since this is real analysis, i think i am mostly interested in the real line. so how do i go about solving it for the reals?
 
  • #4
well we know that the rationals are countable and dense in the reals. but umm..i donno, at this point i am lost about how to prove this.
 
  • #5
That's a good start. Think of all of the intervals (q-1/n,q+1/n) for all q rational and fixed n. That's a countable set of intervals, right? Pick one x in your infinite set in each interval (if there is one). That's a countable set of x's, correct? Now union all of those countable sets over n. Isn't that dense? There's more economical ways to do this, but who cares?
 
  • #6
i think i get what u're saying.. so for every q rational for a fixed n we choose and x in the set that belongs to (q-1/n ; q+1/n). Giving us a set of countable x's, then we do this for all n, which is a countable number of countable sets, thus countable.

But i am having a little trouble seein that this is dense. To show this we need that any point in the infinite set is either in the union of our new sets or a limit point.
To show that any point will be a limit point we need that there is a point within 1/n of it for every n. Certainly there is a rational number 1/n close to it, but are there x's in that range too?

Oh wait a minute...i think i missunderstood the phrase you said "(if there is one)" .. There is definitely one, since there are infinite points and only countable intervals (pigeonhole). (unless by infinite we mean countable, then we definitely have a countable dense subset, since the set is dense in itself). Unless (q - 1/n; q+1/n) is totally disjoint from the set. I just need some convincing that every point is now a limit point
 
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  • #7
Any given interval may miss hitting your set. That's not a problem. Call the countable subset of X that we have defined A. Now pick any point x in X. x is either isolated or it's a limit point of X. If it's isolated then you can find one of the rational intervals that contains only x. By the way we have constructed A, that means x must be in A. Right? Now suppose it's a limit point. Pick any epsilon. Then there is a y in X at a distance less than epsilon from x. Find a rational interval that contains y and fits inside of the epsilon interval. By the construction of A, either y is in A or there is a z in A in the rational interval. In either case there is a point from A within epsilon of x.
 

What does it mean for a set to be "infinite"?

An infinite set is one that has an infinite number of elements, meaning that the set cannot be counted or listed in its entirety. This is in contrast to finite sets, which have a limited number of elements that can be counted.

What is a "countable" set?

A countable set is one that can be put into a one-to-one correspondence with the set of natural numbers (1, 2, 3, ...). This means that each element in the set can be assigned a unique natural number, and there are no elements left out.

What does it mean for a set to be "dense"?

A dense set is one in which every point in the set is arbitrarily close to at least one other point in the set. In other words, there are no "gaps" or "holes" in the set, and every point can be reached by moving a finite distance from any other point in the set.

Why is it important for an infinite set to have a countable dense subset?

This property ensures that the infinite set has enough "structure" to be able to be studied and analyzed in a meaningful way. By having a countable dense subset, the set becomes more manageable and allows for further exploration and understanding.

Can every infinite set have a countable dense subset?

Yes, it is a fundamental property of infinite sets that they can be broken down into smaller, more manageable subsets. This is known as the "countable infinity" property, and it applies to all infinite sets, regardless of their size or complexity.

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