Proof of A Dense in Rn Not Bounded in Math

In summary, the conversation discusses the relationship between a set A being dense and bounded in the space of real numbers. It is stated that if A is dense, then it cannot be bounded, and the conversation explores how to prove this statement. The use of proof by contradiction and the concept of the contrapositive is suggested as a way to prove this statement without making unnecessary assumptions.
  • #1
ELESSAR TELKONT
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I have the following [itex]A\subset\mathbb{R}^{n}[/itex] is dense then [itex]A[/itex] isn't bounded. Is this true? I know that [itex]A[/itex] is dense iff [itex]\bar{A}=\mathbb{R}^{n}[/itex] and that [itex]A[/itex] is bounded iff [itex]\exists \epsilon>0\mid B_{\epsilon}(0)\supset A[/itex]. How to proof it? Or there is an counterexample?
 
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  • #2
Try reductio ad absurdum (proof by contradiction).
 
  • #3
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
matt grime said:
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".
 
  • #5
What? The point was just make a constructive proof of the contrapositive statement without making an unnecessary preliminary assumption.
 
  • #6
matt grime said:
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.
 
  • #7
Thank you for the help!
 

FAQ: Proof of A Dense in Rn Not Bounded in Math

What is the definition of "A Dense in Rn Not Bounded" in mathematics?

A set A is considered dense in Rn if every point in Rn can be approximated by a point in A. A set is considered unbounded if it has no finite bound or limit. Therefore, "A Dense in Rn Not Bounded" means a set in Rn that contains points that can approximate any point in Rn, but has no finite bound or limit.

What is the importance of studying "A Dense in Rn Not Bounded" in mathematics?

Studying "A Dense in Rn Not Bounded" is important because it helps us understand the properties and behavior of unbounded sets in mathematical spaces. It also has applications in various fields such as topology, analysis, and geometry.

How is "A Dense in Rn Not Bounded" different from "A Dense in Rn"?

The main difference between "A Dense in Rn Not Bounded" and "A Dense in Rn" is that the former does not have a finite bound or limit, while the latter may have a finite bound or limit. In other words, "A Dense in Rn Not Bounded" contains points that are infinitely far apart, while "A Dense in Rn" may contain points that are only a finite distance apart.

Can a set be both dense and bounded in Rn?

Yes, it is possible for a set to be both dense and bounded in Rn. For example, the set of rational numbers between 0 and 1 is both dense (every real number between 0 and 1 can be approximated by a rational number) and bounded (the set has a finite limit, 1).

What are some real-world examples of "A Dense in Rn Not Bounded"?

Some real-world examples of "A Dense in Rn Not Bounded" include the set of all real numbers (which is unbounded) and the set of all irrational numbers (which is dense in the set of real numbers). These sets have applications in various fields such as physics, engineering, and economics.

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