# Linear Mapping in Subsets

Question 1.

1) Suppose (V, | * |) is a normed space. If x, y E V and r is a positive real number, show that the open r-balls Br(x) and Br(x + y) in V are homeomorphic.

2) Suppose V and W are two normed spaces. If A : V ---> W is a linear map, then show that it is continuous at every point v E V if and only if it is continuous at 0 E V.

3) Suppose A: (V, | * |) ---> (W, | * |w) is a linear map between normed spaces, and there is a number R E R such that |A(v)|w <= R|v|_v for all v E V. Explain why A is continuous.

Question 2

Let (0,1) denote the open unit interval in R, and C(0,1) the set of all continuous functions (0,1) ---> R. Is C(0,1) a subset of B((0,1),R) the set of all bounded functions on (0,1)? Is C(0,1) a normed space with the sup-norm | * | given by |f| = sup_(t E (0,1){|f(t)|}?

Question 3

1) For a compact topological space, and Y a compact subset of X. The inclusion i: Y --->X gives a function i* : B(X,R) ---> B(Y,R), from the bounded functions on X to the bounded functions on Y by i*(f) = f in i for each f E B(X,R). Explain why i* is subjective.

2) Using the sup norm | * | on both these sets of bounded functions, for a function f E B(X,R), what, is any, is the relation between |f| and |i* (f)|. Is i* continuous?

3) For a compact topological space X, denoted by C(X) the banach space of continuous functions on X with the usual sup-norm.Following the idea of part (1), explain why a continuous function alpha: X --->Z between two compact topological spaces gives a function alpha*:C(Z) ---> C(X)

4) Explain why alpha* of part (3) is a linear map

5) Explain why alpha* of part (3) is continuous.

6) If I: X ---> X is the identity function, show that I*, as in part (3), is the identity function C(X) ---> C(X).

7) If alpha: X_1 ---> X_2 and beta: X_2 ---> X_3 are continuous functions of compact topological spaces, explain why (beta is in alpha)* = alpha* is in beta*

8) Hence prove that if gama: X --> Z is a homeomorphism of compact topological spaces, gama*: C(Z) ---> C(X) is a homeomorphism.

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#### Solution Preview

Please see attached Document for detailed solution.

Problem #1

1. Proof:

We consider the map, such that for any, we have. Since, then we have

Thus . The map is well-defined.

Then I show that is continuous. We consider any and , we can find some , such that . Then for any , we have , then we have

Therefore, is continuous.

Similarly, we can define with , then we can also prove that is continuous.

Therefore, and are homeomorphism.

2. Proof:

From 1, we know that and are homeomorphism, then we have a continuous function and is also continuous. Since is ...

#### Solution Summary

Linear mapping in subsets are examined in the solution.

Countable linear ordering isomorphic to subset of rationals

Show that every countable linear ordering is isomorphic to some subset of the rationals under their usual order, but that omega_1 (the least uncountable ordinal) with its well order is not isomorphic to any set of reals under their usual ordering. The solution may use any algebraic facts about the reals.

See attachments for fully formulated problem.

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