# General Topology

1. Aug 29, 2007

### ELESSAR TELKONT

I have the following $A\subset\mathbb{R}^{n}$ is dense then $A$ isn't bounded. Is this true? I know that $A$ is dense iff $\bar{A}=\mathbb{R}^{n}$ and that $A$ is bounded iff $\exists \epsilon>0\mid B_{\epsilon}(0)\supset A$. How to proof it? Or there is an counterexample?

2. Aug 29, 2007

### George Jones

Staff Emeritus
Try reductio ad absurdum (proof by contradiction).

3. Aug 30, 2007

### matt grime

It's slightly nicer to show that a bounded set in R cannot be dense. Many proofs by contradiction are unnecessary - i.e. you wish to show A implies B, so you assume A and not B and show not B implies not A, without any use of the assumption of A.

4. Aug 30, 2007

### HallsofIvy

Staff Emeritus
But you just said "Assume A"!! Anyway, many people would consider proving the contrapositive to be "proof by contradiction".

5. Aug 30, 2007

### matt grime

What? The point was just make a constructive proof of the contrapositive statement without making an unnecessary preliminary assumption.

6. Aug 30, 2007

### George Jones

Staff Emeritus
Yes, I see. As I read the original post, I said to myself "Well, they both can't be true (being dense and bounded)," and this placed proof by contradiction in my mind. Bot all that is needed is ~B (bounded), so contapositive gives a direct proof.

7. Aug 30, 2007

### ELESSAR TELKONT

Thank you for the help!!