I've seen a few different versions of the Bolzano-Weierstrass theorem in some it is specified that the sequence is real and in some it is not specified at all. Does Bolzano-Weierstrass' theorem work for rational sequences, i.e. If a sequence is rational and bounded does it then always have a convergent subsequence?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
It has a subsequence which converges to a real number but not necessarily to a rational number
For example $3, \frac{31}{10}, \frac{314}{100}, \frac{3141}{1000}, \ldots$ is the start of a bounded rational sequence with every subsequence converging to $\pi$, which is not rational
