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I know that a sequence can have many different subsequential limits but is the number of subsequential limits always countable? How do we know?

mathjacks
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1 Answers1

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Consider the sequence $0,1/2,1/3,2/3,1,1/4,2/4,3/4,1,\cdots$. It's subsequences can converge to any irrational number in $[0,1]$. So definitely not countable.

Proof: Any irrational is the limit of rational numbers $a_n/b_n$, for integers $a_n,b_n$. You can do better: any irrational is a limit of $a'_n/b'_n$ where both $a_n$ and $b_n$ are strictly increasing (multiply the $a_n,b_n$ by a large enough constant $c_n$, this doesn't change their ratio). Since $b_n'$ is increasing, you can now pull out a subsequence from the above.

Alex R.
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