Question 3.T.12: (Bolzano-Weierstrass) Every infinite and bounded subset of R...
(Bolzano-Weierstrass)
Every infinite and bounded subset of \mathbb{R} has at least one cluster point in \mathbb{R}.
Let A be an infinite and bounded set of real numbers. Being bounded, A is contained in some bounded closed interval I_{0} = [a_{0}, b_{0}].
First we bisect I_{0} into the two subintervals
I^{′}_{0} = \left[a_{0},\frac{a_{0} + b_{0}}{2} \right] , I^{″}_{0} = \left[\frac{a_{0} + b_{0}}{2},b_{0}\right].Since A ⊆ I^{′}_{0} ∪ I^{″}_{0} is infinite at least one of the sets A ∩ I^{′}_{0} and A ∩ I^{″}_{0} is infinite. Let I_{1} = [a_{1} , b_{1}] = I^{′}_{0} if A ∩I^{′}_{0} is infinite, otherwise set I_{1} = I^{″}_{0}.
We then have
(i) I_{1} ⊆ I_{0}
(ii) l(I_{1}) = \frac{1}{2} (b_{0} − a_{0})
(iii) A ∩ I_{1} is infinite.
Now we bisect I_{1} into the subintervals
I^{′}_{1} = \left[a_{1},\frac{a_{1} + b_{1}}{2} \right] , I^{″}_{1} = \left[\frac{a_{1} + b_{1}}{2},b_{1}\right],one of which necessarily intersects A in an infinite set. Define I_{2} = [a_{2}, b_{2}] to be I^{′}_{1} if A ∩ I^{′}_{1} is infinite, otherwise let I_{2} =I^{″}_{1}. Continuing in this fashion, we obtain, in the n-th step, the intervals I_{i} = [a_{i}, b_{i}], 0 ≤ i ≤ n, which satisfy
(i) I_{i} ⊆ I_{i−1}
(ii) l(I_{i}) = \frac{1}{2^{i}}(b_{0} − a_{0})
(iii) A ∩ I_{i} is infinite.
Let us again bisect I_{n} to obtain
I^{′}_{n} = \left[a_{n},\frac{a_{n} + b_{n}}{2} \right] , I^{″}_{n} = \left[\frac{a_{n} + b_{n}}{2},b_{n}\right].Since I_{n} = I^{′}_{n} ∪ I^{″}_{n} and A ∩ I_{n} is infinite, either A ∩ I^{′}_{n} is infinite, in which case we set I_{n+1} = [a_{n+1}, b_{n+1}] = I^{′}_{n} , or A∩I^{′}_{n} is finite and A∩I^{″}_{n} is infinite, in which case we choose I_{n+1} = I^{″}_{n}. Now we see that
(i) I_{n+1} ⊆ I_{n}
(ii) l(I_{n+1}) = \frac{1}{2} l(I_{n}) = \frac{1}{2^{n+1}}(b_{0} − a_{0})
(iii) A ∩ I_{n+1} is infinite.
By induction we therefore conclude the existence of a sequence (I_{n} : n ∈ \mathbb{N}) of closed, bounded intervals that satisfies conditions (i), (ii), and (iii) above. By Cantor’s theorem there is a (single) point x ∈ \cap ^{\infty }_{n=1} I_{n}.
We complete our proof by showing that x ∈ Â.
Given any ε > 0 we can choose n ∈ \mathbb{N} so that
\frac{b_{0} − a_{0}}{2^{n}} < ε,or, equivalently, |I_{n}| < ε. This, together with the fact that x ∈ I_{n} for all n, implies
I_{n} ⊂ (x − ε, x + ε).Since I_{n} contains an infinite number of elements of A, so does (x − ε, x + ε), hence x ∈ Â.