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Measure, Integration & Real Analysis, 2021a

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186 Chapter 6 Banach Spaces<br />

Because<br />

R = ⋃<br />

{x}<br />

x∈R<br />

and each set {x} has empty interior in R, Baire’s Theorem implies R is uncountable.<br />

Thus we have yet another proof that R is uncountable, different than Cantor’s original<br />

diagonal proof and different from the proof via measure theory (see 2.17).<br />

The next result is another nice consequence of Baire’s Theorem.<br />

6.80 the set of irrational numbers is not a countable union of closed sets<br />

There does not exist a countable collection of closed subsets of R whose union<br />

equals R \ Q.<br />

Proof This will be a proof by contradiction. Suppose F 1 , F 2 ,... is a countable<br />

collection of closed subsets of R whose union equals R \ Q. Thus each F k contains<br />

no rational numbers, which implies that each F k has empty interior. Now<br />

( ⋃ ) ( ⋃ ∞<br />

R = {r} ∪ F k<br />

).<br />

r∈Q<br />

k=1<br />

The equation above writes the complete metric space R as a countable union of<br />

closed sets with empty interior, which contradicts Baire’s Theorem [6.76(a)]. This<br />

contradiction completes the proof.<br />

Open Mapping Theorem and Inverse Mapping Theorem<br />

The next result shows that a surjective bounded linear map from one Banach space<br />

onto another Banach space maps open sets to open sets. As shown in Exercises 10<br />

and 11, this result can fail if the hypothesis that both spaces are Banach spaces is<br />

weakened to allow either of the spaces to be a normed vector space.<br />

6.81 Open Mapping Theorem<br />

Suppose V and W are Banach spaces and T is a bounded linear map of V onto W.<br />

Then T(G) is an open subset of W for every open subset G of V.<br />

Proof Let B denote the open unit ball B(0, 1) ={ f ∈ V : ‖ f ‖ < 1} of V. For any<br />

open ball B( f , a) in V, the linearity of T implies that<br />

T ( B( f , a) ) = Tf + aT(B).<br />

Suppose G is an open subset of V. Iff ∈ G, then there exists a > 0 such that<br />

B( f , a) ⊂ G. If we can show that 0 ∈ int T(B), then the equation above shows that<br />

Tf ∈ int T ( B( f , a) ) . This would imply that T(G) is an open subset of W. Thus to<br />

complete the proof we need only show that T(B) contains some open ball centered<br />

at 0.<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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