Since the last lecture covered some difficult material and its presentation in the book [Chapter 5, Section 5] is very sketchy, I have decided to write-up the results in more detail. The same is true for the today’s lecture [see Chapter 5, Section 7].

1) Let X be a subset of a normed vector space and {x_{n}}, n > 0 be sequence of points in X. We say that a sequence {y_{m}}, m > 0 of points in X is a subsequence of {x_{n}} if there exists a increasing sequence n_{1} < n_{2} < ... < n_{m} < ... of integers such that y_{m} = x_{n_{m}}, m > 0. [So y_{1} = x_{n_{1}} ...].

Lemma. If {y_{m}} ∈ X, m > 0 is a subsequence of {x_{n}} and {z_{r}}, r > 0 is a subsequence of {y_{m}} then {z_{r}}, r > 0 is a subsequence of {x_{n}}.

Proof of Lemma.

Since {y_{m}}, m > 0 is a subsequence of {x_{n}} we can find an increasing sequence of n_{1} < n_{2} < ... < n_{m} < ... of integers such that y_{m} = x_{n_{m}}, m > 0. Since {z_{r}}, r > 0 is a subsequence of {y_{m}} we can find an increasing sequence of m_{1} < m_{2} < ... < m_{r} < ... of integers such that z_{r} = y_{m_{r}}, r > 0. Consider now the sequence ñ_{r} of integers where ñ_{r} := n_{m_{r}}. It is clear that ñ_{r} is an increasing sequence of integers and z_{r} = xñ_{r}, r > 0. The lemma is proved.

2)Let (V, ∥v∥_{V}) be a normed vector space, X ⊂ V. We say that X is compact if for any sequence {x_{n}} ∈ X, n > 0 we can find a subsequence {y_{m}} of {x_{n}} and a point x_{0} ∈ X such that the subsequence {y_{m}} is convergent to x_{0}.

3)Let (V′, ∥v′∥_{V′}), (V′′, ∥v′′∥_{V′′}) be normed vector spaces. Let V be the direct product V′ × V′′ [see Chapter 2, Section 2 Problem 7 in Corwin, Szczarba]. We denote by ∥v∥_{max} the norm on V such that ∥(v′, v′′)∥_{max} = max(∥v′∥_{V′}, ∥v′′∥_{V′′}).

Check that (V, ∥v∥_{max}) is a normed vector space!

For any subsets X′ ⊂ V′, X′′ ⊂ V′′ define a subset X := X′ × X′′ ⊂ V′ × V′′ as the set of pairs x = (x′, x′′) where x′ ∈ X, x′′ ∈ X′′.

Theorem 1. For any pair of compact subsets X′ ⊂′, X′′ ⊂ V′′ the set X := X′ × X′′ is compact.

Proof. Let {x_{n}} ∈ X, n > 0 be any sequence. We can write x_{n} = (x′_{n}, x′′_{n}), x′_{n} ∈ X, x′′_{n} ∈ X′′. Since the set X′ is compact there exists an increasing sequence n_{1} < n_{2} < ... < n_{m} < ... of integers and a point x′_{0} ∈ X′ such that the subsequence y′_{m} := x′_{n_{m}}, m > 0 is convergent to x′_{0}.

Consider now the sequence {y_{m}} ∈ X, y_{m} := x_{n_{m}}, m > 0. Then {y_{m}} is a subsequence of the sequence {x_{n}}, y_{m} = (y′_{m}, y′′_{m}), y′_{m} ∈ X′, y′′_{m} ∈ X′′ and the sequence {y′_{m}} is convergent to x′_{0} ∈ X′.

Consider now the sequence {y′′_{m}}, m > 0. Since the set X′′ is compact there exists an increasing sequence m_{1} < m_{2} < ... < m_{r} < ... of integers and a point x′′_{0} ∈ X′′ such that the subsequence {z′′_{r}}, z′′_{r} := y′′_{m_{r}}, r > 0 of the sequence {y′′_{m}} is convergent to x′′_{0}. As follows from the proof of Lemma there exists an increasing sequence ñ_{r}, r > 0 of integers such that z′′_{r} := x′′ ; ñ_{r}, r > 0. Consider the sequence {z_{r}}, z_{r} := xñ_{r} ∈ X, r > 0. Then z_{r} = (z′_{r}, z′′_{r}), z′_{r} ∈ X′, z′′_{r} ∈ X′”. By the construction the sequence z′′_{r} is convergent to x′′_{0} ∈ X′′. On the other hand, by the construction the sequence z′_{r} is a subsequence of a convergent sequence y′_{m}. Therefore the sequence z′_{r} is is convergent to x′_{0} ∈ X′. So the sequence {z_{r}} is is convergent to x_{0} := (x′_{0}, x′′_{0}) ∈ X in (V, ∥v∥_{max}). Since, by the construction the sequence {z_{r}} is a subsequence of the sequence {x_{n}} we see that any sequence {x_{n}} ∈ X, n > 0 has a convergent subsequence. Theorem 1 is proved.

Theorem 2. Let f be a continuous function on a closed interval [a, b] such that f(a) < 0, f(b) > 0. Then there exists a point c ∈ [a, b] such that f(c) = 0.

Proof of the Theorem 2. Let L ⊂ ℝ be the half-line L = {y ∈ ℝ|y ≤ 0} and X := f^{−1}(L) ⊂ [a, b] be the preimage of Y (that is X = {x ∈ [a, b]|f(x) ≤ 0}). Since f is continuous and Y is closed we know (?) that the subset X ⊂ [a, b] is closed. Since f(a) < 0 we see that a ∈ X and therefore X is not empty. Since X ⊂ [a, b] we know that X is bounded. Therefore we can define c := lub(X) (where lub denotes the least upper bound). Since X is closed we know (?) that c ∈ X. Therefore f(c) ≤ 0. I claim that f(c) = 0. To prove that f(c) = 0 we show that the assumption that f(c) ≠ 0 leads to a contradiction.

So assume that f(c) ≠ 0. Since we know that f(c) ≤ 0 the assumption that f(c) ≠ 0 implies that f(c) < 0. So f(c) = −ϵ, ϵ > 0. Since f is continuous we can find δ > 0 such that for all x ∈ [a, b], |x − c| < δ we have |f(x) − f(c)| < ϵ. In particular, |f(c + δ/2) − f(c)| < ϵ and therefore f(c + δ/2) < f(c) + ϵ = 0. So c + δ/2 ∈ X. But we have defined c as the least upper bound of X. CONTRADICTION. Theorem 2 is proved.