Proof of A Dense in Rn Not Bounded in Math

[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?

 

[George Jones]

Try reductio ad absurdum (proof by contradiction).

 

[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.

 

[HallsofIvy]

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.

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

 

[matt grime]

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

 

[George Jones]

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

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.

 

[ELESSAR TELKONT]

Thank you for the help!