Does Every Point of R Exist as an Accumulation Point in a Sequence?

AI Thread Summary
A sequence can be constructed where every point in the real numbers R is an accumulation point, particularly using the set of rational numbers, which are dense in R. The discussion highlights that rational numbers can be enumerated due to their countability, supporting the idea that they can serve as accumulation points for every real number. Additionally, the Heine-Borel Theorem is illustrated by showing that a non-closed bounded set, such as the open interval (-1, 1), can be covered by an open cover that lacks a finite sub-cover. This leads to a contradiction, emphasizing the necessity of closed sets in the theorem. Understanding these concepts is crucial for deeper insights into real analysis.
Kraziethuy
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1.Does a sequence exist that has every point of R(real numbers) as an accumulation point?

2.Show that closed is essential in the Heine-Borel Theorem by finding an open cover of a non-closed bounded set that does not have a finite sub-cover.

I think that the set of rational numbers has every real numbers as an accumulation point, but I'm unsure of the sequence of rational numbers.

Any help appreciated, thanks.
 
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I think that the set of rational numbers has every real numbers as an accumulation point, but I'm unsure of the sequence of rational numbers.

You have the right idea. A potentially unsatisfying answer might be to simply enumerate the rational numbers (which is possible because they are countable).
Alternatively, are you familiar with Hilbert's hotel and the proof that the rational numbers are countable?
 
For Q2, just consider the open interval of (-1, 1) in R. Consider an open cover like the collection of intervals, { (x-w, x+w), w=(1-|x|)/2: x from (-1, 1) }. If this collection has a finite subcover, would there be any points not covered by the subcovering (and thus a contradiction)?
 
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