# Math Help - Infinite sets

1. ## Infinite sets

Prove that if A = R in (R; Euclidean metric) then the set A must be infinite.
Where A is a closed set.

Surely A is Infinite because it's equal to an infinite set.

How will you prove this ???

2. Indeed, I don't see the point. Maybe the question is : every dense subset of $\mathbb R$ must be infinite.

3. As girdav suggests...maybe the problem is asking about the closure of A in R?

4. Sorry I made a mistake, this is the question,

Prove that if A = R in (R; Euclidean metric) then the set A' must be infinite.
Where A is a closed set.
and A' is open.

Wouldn't A, be the empty set, How can it be infinite???

5. Originally Posted by Dreamer78692
Sorry I made a mistake, this is the question,
Prove that if A = R in (R; Euclidean metric) then the set A' must be infinite. Where A is a closed set. and A' is open.
Even after the correction in blue above, the question still makes no sense. Are you doing a translation of the original source?

The bit in red above worries me most. If right, it seems to trivialize the question.

As for your correction, the notation $A^{\prime}$ as several different meanings in mathematics. In topology it is widely used to denote the set of all limit points of $A$, the derived set.
In some set theory textbooks the notation means the complement of the set.
I think that you must post what your text material says about $A^{\prime}$. How is it used?

Also, please review the exact wording of this problem.

6. Please See attachment for Question

7. Sorry, but that does not help. In fact it contradicts your own correction.
Please post the entire question from start to finish, not just part c.
Also answer my question about the way the textbook defines $A^{\prime}$.

8. This is a True and False Question, Never copy the question properly... SORRY!!!

9. Okay, so that is what tinyboss suggested initially: the [b]closure[b] of A, not A itself, is equal to R- A is dense in R. I think the simplest way to prove that a set dense in R must be infinite is an indirect proof: If A were finite, then there would be a largest member, m. Then look at m+1 to show that this contradicts the fact that A is dense in R.