Two fixed point theorems in Banach space

Here is a famous fixed point theorem in finite dimension by Brouwer:

Theorem 1 (Brouwer fixed point theorem) Let {M \subset {\mathbb R}^n} be a convex compact set, for any continuous function {f : M \rightarrow M}, there exists a point {x_0 \in M} such that {f (x_0) = x_0}.

There are couple of ways to extend this theorem to Banach spaces. First recall that a mapping between two Banach spaces is called compact, if the mapping is continuous (not necessarily linear) and the images of bounded sets are pre-compact.

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Equivalence of compactness in metric space

Lemma In a metric space {(X,d)}, if {\{C_k\}_{k=1}^\infty \subset X} is a sequence of compact sets such that {C_{k+1} \varsubsetneq C_k} for {k \in {\mathbb N}} and {\lim_{k \rightarrow \infty}} diam{C_k = 0}. Then {\cap_{k=1}^\infty C_k = \{c\}}, where {c} is a point in {X}.

Proof: First note that {\cap_{k=1}^\infty C_k} cannot contain more than 1 points. If {a,b \in \cap_{k=1}^\infty C_k}, then {a,b \in C_k \quad \forall k}. Since {\lim} diam{C_k}=0, it forces {a=b}.

It remains to show {\cap_{k=1}^\infty C_k} is nonempty. Assume it is empty, then {\cup_{k=1}^\infty (X \setminus C_k ) = X \setminus (\cap_{k=1}^\infty C_k) = X}. Since metric space is Hausdorff, {C_k} is closed for all {k} and {\{ X \setminus C_k \}} is an open cover for {C_1}. Since {C_1} is compact, there exists finite subcover such that {C_1 \subset \cup_{i=1}^m (X\setminus C_{k_i}) = X \setminus C_{k_m}}, which contradicts {C_{k_m} \varsubsetneq C_1}. \Box

Theorem In a metric space {(X,d)} (no matter it has countable basis or not), let {E} be a subset in {X}. The following statement are equivalent by assuming axiom of choice:

  1. {E} is compact;
  2. {E} is sequentially compact;
  3. {E} is complete and totally bounded.

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