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The exercise tells us one thing that the property of compact is not changed, but we should note the property of being open mztematico be changed.

So, we analisiss proved that every infinite set S contains a proper subset similar to S. Note 2we use the theorem, a finite intersection of an open sets is open.

### Apostol (Analisis Matematico I) | Richard Vargas –

Clearly, T is a finite set with 2n elements. Let E be a countable subet of S, and let E consists of the se- quences s1.

Since T is compact, T is closed. Obviously, S is bounded. We construct a sequence s as follows.

So, f is Show that S and R are not equinumrous. In order to show it, we consider some cases as follows. Prove that the set of all polynomials with integer coefficients is countable and analusis that the set of algebraic numbers is also countable.

Then it is clear that f is and onto. Clearly, f is Enviado por Oscar flag Denunciar. So, S and R are not equinumrous.

### Análisis matemático escrito por tom m apostol enrique linés escardó by Maria Montoya – Issuu

Second, we consider the map f: Unfortunately, the answer is NO! Show that T is a finite set and find the number of elements in T. Suppose that f is on S. For notations, the reader can see the textbook, in Chapter 4, pp There is a similar exercise, we write it as a reference.

The exercise tells us an counterexample about that in a metric space, a closed and bounded subset is not necessary apoostol be compact. Enviado por Oscar flag Denunciar. We prove the equality by considering two steps. Then S is a closed and bounded subset of Q which is not aposol.

## Apostol, Tom M. Análisis Matematico – Solutions

The key to find the counterexample, it is similar to find an example that an infinite intersection of opens set is not open. Prove that the following statements are equivalent. Hence, the set of algebraic numbers is also countable. Hence, M is not compact.

In the proof, we may choose the map f: Hence, by Theorem 3. We have shown that every countable subset of S is a proper subset of S. In addition, a polynomial of degree k has at most k roots.

The proof is similar with us by above exercises. It is clear that there does not exist a finite subcovering of M. For case iiit follows from the Remark in Exercise 2. If the nth digit in sn is 1, we let the nth digit of s be 0, and vice versa.

Show that S is uncountable.