## 3.3. Subsequences

### Theorem 3.3.4: Bolzano-Weierstrass

Let be a sequence
of real numbers that is bounded. Then there exists a subsequence
that converges.

### Proof:

Since the sequence is bounded, there exists a number*M*such that

*| a*for all

_{j}| < M*j*. Then:

eitherSay that[-M, 0]or[0, M]contains infinitely many elements of the sequence

*[0, M]*does. Choose one of them, and call it

eitherSay it is[0, M/2]or[M/2, M]contains infinitely many elements of the (original) sequence.

*[0, M/2]*. Choose one of them, and call it

eitherThis time, say it is[0, M/4]or[M/4, M/2]contains infinitely many elements of the (original) sequence

*[M/4, M/2]*. Pick one of them and call it

Keep on going in this way, halving each interval from the previous step at the next step, and choosing one element from that new interval. Here is what we get:

*| - | < M*, because both are in*[0, M]**| - | < M / 2*, because both are in*[0, M/2]**| - | < M / 4*, because both are in*[M/2, M/4]*

because both are in an interval of length| - | < M / 2^{k-1}

*M / 2*. So, this proves that consecutive elements of this subsequence are close together. That is not enough, however, to say that the sequence is Cauchy, since for that not only consecutive elements must be close together, but all elements must get close to each other eventually.

^{k-1}
So: take any * > 0*, and
pick an integer *N* such that
???...??? (This trick is often used: first, do some calculation, then
decide what the best choice for *N* should be. Right now, we
have no way of knowing a good choice). Pretending, however, that we knew
this choice of *N*, we continue the proof. For any
*k, m > N* (with *m > k*) we have:

Now we can see the choice for

*N*: we want to make is so large, such that whenever

*k, m > N*, the difference between the members of the subsequence is less than the prescribed . What is therefore the right choice for

*N*to finish the proof ?