The Bolzano–Weierstrass theorem in real analysis says that every bounded sequence of real numbers contains a convergent subsequence.

The sequence a1, a2, a3, ... is called bounded if there exists a number L that the absolute value |an| is less than L for every "n". This can be imagined as points "ai" on a graph. With "i" on the horizontal axis and the value on the vertical. The sequence then travels to the right as it progresses, and it is bounded if we can draw a horizontal strip which has all of the points.

A subsequence is a sequence that omits some members, for example a2, a5, a13, blah blah blah so on...

Here is somwwhat of a proof:

Start with a FINITE interval that contains all the "an". Since the sequence is bounded,( -L, L )cut it into halves. At least one half must contain "an" for infinitely many n. and then continue with that half and cut it into two halves, so on and so forth. This process constructs a sequence of intervals where the common element is the limit of a subsequence.

Since every subsequence of a convergent sequence comes together, the proof can be generalized to bounded sequences in "Rn" by using one component at a time.

I tried to break it down to make sense out of it but pretty much confused myself a bit. This could be totally wrong but nonetheless tried. One thing about mathematics, I've always had a hard time with is that its harder to explain in words than to just sit there and do it.

Is this wrong Phlog?