Some examples of Banach spaces that have been discussed up to now are ℝ^{n}, ℂ^{n}, and L^{p}
The following remarkable result is called the Baire category theorem. To get an idea of its meaning, imagine you draw a line in the plane. The complement of this line is an open set and is dense because every point, even those on the line, are limit points of this open set. Now draw another line. The complement of the two lines is still open and dense. Keep drawing lines and looking at the complements of the union of these lines. You always have an open set which is dense. Now what if there were countably many lines? The Baire category theorem implies the complement of the union of these lines is dense. In particular it is nonempty. Thus you cannot write the plane as a countable union of lines. This is a rather rough description of this very important theorem. The precise statement and proof follow.
Theorem 23.1.2 Let (X,d) be a complete metric space and let {U_{n}}_{n=1}^{∞} be a sequence of open subsets of X satisfying U_{n} = X(U_{n} is dense). Then D ≡∩_{n=1}^{∞}U_{n} is a dense subset of X.
Proof: Let p ∈ X and let r_{0} > 0. I need to show D ∩B(p,r_{0})≠∅. Since U_{1} is dense, there exists p_{1} ∈ U_{1} ∩ B(p,r_{0}), an open set. Let p_{1} ∈ B(p_{1},r_{1}) ⊆B(p_{1},r_{1}) ⊆ U_{1} ∩ B(p,r_{0}) and r_{1} < 2^{−1}. This is possible because U_{1} ∩ B

because


and let r_{2} < 2^{−2}. Continue in this way. Thus r_{n} < 2^{−n},


The sequence, {p_{n}} is a Cauchy sequence because all terms of

Since all but finitely many terms of {p_{n}} are in B(p_{m},r_{m}), it follows that p_{∞}∈B(p_{m},r_{m}) for each m. Therefore,

The following corollary is also called the Baire category theorem.
Corollary 23.1.3 Let X be a complete metric space and suppose X = ∪_{i=1}^{∞}F_{i} where each F_{i} is a closed set. Then for some i, interior F_{i}≠∅.
Proof: If all F_{i} has empty interior, then F_{i}^{C} would be a dense open set. Therefore, from Theorem 23.1.2, it would follow that

The set D of Theorem 23.1.2 is called a G_{δ} set because it is the countable intersection of open sets. Thus D is a dense G_{δ} set.
Recall that a norm satisfies:
a.)
b.)
c.)
From the definition of continuity, it follows easily that a function is continuous if

implies

Theorem 23.1.4 Let X and Y be two normed linear spaces and let L : X → Y be linear (L(ax + by) = aL(x) + bL(y) for a,b scalars and x,y ∈ X). The following are equivalent
a.) L is continuous at 0
b.) L is continuous
c.) There exists K > 0 such that
Proof: a.)⇒b.) Let x_{n} → x. It is necessary to show that Lx_{n} → Lx. But (x_{n} − x) → 0 and so from continuity at 0, it follows

so Lx_{n} → Lx. This shows a.) implies b.).
b.)⇒c.) Since L is continuous, L is continuous at 0. Hence Lx_{Y } < 1 whenever x_{X} ≤ δ for some δ. Therefore, suppressing the subscript on the ,

Hence

c.)⇒a.) follows from the inequality given in c.). ■
Definition 23.1.5 Let L : X → Y be linear and continuous where X and Y are normed linear spaces. Denote the set of all such continuous linear maps by ℒ(X,Y ) and define
 (23.1) 
Note that from Theorem 23.1.4
The next lemma follows immediately from the definition of the norm and the assumption that L is linear.
Proof: Let x≠0 then x∕

Therefore, multiplying both sides by

It remains to verify the triangle inequality. Let L,M ∈ℒ
For example, consider the space of linear transformations defined on ℝ^{n} having values in ℝ^{m}. The fact the transformation is linear automatically imparts continuity to it. You should give a proof of this fact. Recall that every such linear transformation can be realized in terms of matrix multiplication.
Thus, in finite dimensions the algebraic condition that an operator is linear is sufficient to imply the topological condition that the operator is continuous. The situation is not so simple in infinite dimensional spaces such as C
Proof: Let {L_{n}} be a Cauchy sequence in ℒ(X,Y ) and let x ∈ X.

Thus {L_{n}x} is a Cauchy sequence. Let

Then, clearly, L is linear because if x_{1},x_{2} are in X, and a,b are scalars, then
