Density and Sequences, Real Analysis

[onemachine]

Suppose a set containing a sequence {a_n} is dense in an interval of real numbers [a, b].

Then must it follow that the sequence a_n is onto said interval [a, b]?

 

[SlipEternal]

Suppose a set containing a sequence {a_n} is dense in an interval of real numbers [a, b].

Then must it follow that the sequence a_n is onto said interval [a, b]?

Assume that the sequence \(\displaystyle \{a_n\}\) is the image of the function: \(\displaystyle f:\mathbb{N}\to [a,b]\). Let \(\displaystyle f(n)=a_n\). \(\displaystyle [a,b]\subset \mathbb{R}\) is uncountable. \(\displaystyle \{a_n\}\) is countable. So, use Cantor's diagonal argument to show that the sequence is not onto.

 

[onemachine]

\(\displaystyle \{a_n\}\) is countable.

If \(\displaystyle \{a_n\}\) is dense in a real interval, doesn't that imply that \(\displaystyle \{a_n\}\) is uncountable.

 

[SlipEternal]

The rationals are dense in the reals. The rationals are countable. So, there are most definitely countable dense subsets of uncountable sets.

Edit: And, if you don't believe me, and you think that \(\displaystyle \{a_n\}\) is onto, then do what I said. If \(\displaystyle \{a_n\} = [a,b]\), then construct Cantor's diagonal argument, and you will see that \(\displaystyle [a,b]\) has more elements than \(\displaystyle \{a_n\}\).

 

[onemachine]

Subquestion: I believe that density does not imply "ontoness" (as I originally asked) otherwise the universe would likely implode due to the unraveling of the spacetime continuum.

But does "ontoness" imply density?

 

[SlipEternal]

Subquestion: I believe that density does not imply "ontoness" (as I originally asked) otherwise the universe would likely implode due to the unraveling of the spacetime continuum.

But does "ontoness" imply density?

If a function is onto, then its image is equal to its range. A set is definitely dense in itself.