1. ## open inertvals (proof)

Hello all,

Problem:

Let A be a bounded subset of $\mathbb{R}$, which contains infinitely many different points. Prove that there exists at least one point $x_{0}$ such that all open intervals around $x_{0}$ contains infinitely many different points from A.

2. This doesn't seem true to me. If $A=\{ \frac{1}{ 2^n}|n\in \mathbb{N}\}$ then the open interval $(\frac{1}{2^n}-\frac{1}{2^{n+1}},\frac{1}{2^n}+\frac{1}{2^{n+1}})$ contains only $\frac{1}{2^n}$.

Is there a flaw in this counterexample I'm not seeing?

3. Originally Posted by DrSteve
This doesn't seem true to me. If $A=\{ 2^n|n\in \mathbb{N}\}$ Is there a flaw in this counterexample I'm not seeing?
In the given, $A$ is a bounded set.

4. Oops - I meant $\frac{1}{2^n}$. I'm going to edit my post. Now tell me if there's something wrong with it.

5. Now that works nicely as a counter-example.
I suspect the OP meant to add that $\math{A}$ is closed.

6. Originally Posted by DrSteve
This doesn't seem true to me. If $A=\{ \frac{1}{ 2^n}|n\in \mathbb{N}\}$ then the open interval $(\frac{1}{2^n}-\frac{1}{2^{n+1}},\frac{1}{2^n}+\frac{1}{2^{n+1}})$ contains only $\frac{1}{2^n}$.

Is there a flaw in this counterexample I'm not seeing?
It doesn't say that the point $x_0$ has to be in A. In this example, take $x_0=0$.

7. Originally Posted by Opalg
It doesn't say that the point $x_0$ has to be in A. In this example, take $x_0=0$.
Oops I missed that too.

8. In this case, I would try to construct a limit point.

First, $A$ can be contained in a closed interval $I=[a,b]$. Split $I$ in half. One of the parts must contain infinitely many points from $A$. Take that one and call it $A_1$. Then, split $A_1$ in half. One of those parts must contain infinitely many points from $A$, so take that one and call it $A_2$. Continue this procedure indefinitely to receive a sequence of closed intervals $A_1, A_2, A_3,\ldots$ such that $A_1\subset A_2\subset A_3,\ldots$ whose sizes are $\frac{b-a}{2},\frac{b-a}{2^2},\frac{b-a}{2^3},\ldots$, respectively.

Now, by the Nested Interval theorem, $\bigcap A_n$ contains a single point, and that point is the limit point we are looking for.

9. Originally Posted by surjective
Hello all,

Problem:

Let A be a bounded subset of $\mathbb{R}$, which contains infinitely many different points. Prove that there exists at least one point $x_{0}$ such that all open intervals around $x_{0}$ contains infinitely many different points from A.
Merely note that since $A$ is bounded we have that $\overline{A}$ is bounded and thus compact. Thus, we have that there is an infinite subset of $\overilne{A}$ and by compactness (or the Bolzano-Weierstrass property to be exact) this sequence must have a limit piont. In any $T_1$ space (if this doesn't mean anything ignore it and replace it with $\mathbb{R}$) a point is a limit point if and only if every neighborhood contains infintely many points of the set.