## Just For Fun (Rudin’s Topology Section)

Just for fun I’ve decided to do the entire section on topology in Rudin’s Principles of Mathematical Analysis. The problems themselves aren’t particularly difficult, but some are interesting and it will give me a chance to do some refreshing. Also, a friend of mine is starting Rudin and it will be nice to have done them in case he has any questions. I will probably prove more general results from topology if I feel like it. So…

**1.**

**Problem: **Prove that the empty set is a subset of every set.

**Proof:** Suppose that we are working in some universal set and there existed some such that . Then, by definition there exists some such that . But, this is clearly preposterous since the statement is *never* true.

**2. **

**Problem: **A complex number is said to be *algebraic* if there are integers , not all zero, such that

Let denote the set of all algebraic numbers. Prove that .

**Proof:** Let . By the fundamental theorem of algebra we have that . So, clearly being the countable union of countable sets is countable. Thus, we claim that

from where it will follows that being a subset of a countable union of countable sets is countable. So, let then there exists such that . Thus, clearly from where the conclusion follows.

**3.**

**Problem:** Prove that there exists real numbers which aren’t algebraic.

**Proof:** Clearly since is countable so is . And thus, if were countable then so would which is of course absurd. It follows that there are *uncountably* many non-algebraic (transcendental) numbers.

**4.**

**Problem:** Is the set of all irrational numbers countable?

**Proof: **Of course not. For using a similar argument to the above, if were countable then would be as well.

**5.**

**Problem:** Construct a bounded set of real numbers with exactly three limit points.

**Proof:** Why not do it for an arbitrary natural number. Let . Evidently has precisely one limit point. Now, let then it is equally clear has exactly one limit point (namely ). So, the set has precisely limit points as one can easily check.

**6.**

**Problem:** Let be the set of all limit points of a set (where is a metric space). Prove that a) is closed , b) , c) Do and always have the same limit points?

**Proof:**

**a) **This is true in any topological space. But, first we need a quick lemma

**Lemma:** Let be a space, and . Then, if then for very neighborhood of we have that is infinite.

**Proof:** Suppose not, and there existed a neighborhood of such that . Since is for each there exists some neighborhood of such that . Thus, is a neighborhood of which does not intersect , which of course contradicts that .

So, using this let be a space, then for any it is true that is closed. To see this let and let be any neighborhood of it. By assumption there exists some and thus since is a neighborhood of also there exists infinitely many points of in . In particular there exists some such that . Since was arbitrary it follows that . The conclusion follows.

** b) **Let then every neighborhood contains a point of different from itself, and thus a point of different from itself. Conversely, let and let be any neighborhood of it. We must have that there is some . Now, if we’re done, and if not by the previous lemma since is a limit point of it must be that contains infinitely many points of anyways, and thus a point of different from . The conclusion follows.

**c) **Of course not. with the usual topology has exactly one limit point, namely . But, a singleton in Euclidean space does not have an limit points.

**7. **

**Problem** Let be a metric space and . a) Show that . b) show that example that and the inclusion can be strict.

**Proof:**

**a) **This is true in any topological space . Suppose then there exists a neighborhood of such that and thus and thus and so . Conversely, let . Then, since there exists some neighborhood of it such that . Thus, clearly is a neighborhood of which does not intersect and thus . The conclusion follows.

**b) **Let then there exists a neighborhood of such that and thus . It follows that and thus . But, as was noted the inclusion can be strict. Give the usual topology. Then, for each we have the singleton is closed and thus . So,

**8.**

**Problem:** a) Is every point of every open set a limit point of ? Answer the same for closed sets in ?

**Proof:**

**a) **Of course, I mean this is true in any topological space. If is open then each point of is an interior point and thus trivially a limit point.

**b) **No, take a singleton.

**9.**

**Problem:** Let denote the set of all interior points for in a metric space . Prove that:

**a) ** is always open

**b) ** is open if and only if .

**c) **If and is open then .

**d) **

**e)** Do and always have the same interiors?

**f) **Do and always have the same interiors?

**Proof:**

**a) **Let then there exists a neighborhood of such that , but for each we also have that is a neighborhood of it which is contained in . Thus, .

**b) **If the above proves that is open. Conversely, since a set in a metric space is open if and only if each point is an interior point the converse is trivial.

**c) **Let then is automatically a neighborhood of which is contained in so that .

**d) **Let then for every neighborhood of we have that and so and thus . Conversely, let then for every neighborhood of we have that and thus and so or .

**e) **Of course not. It is trivial to check that if one gives the usual topology then (since it’s complement is dense in ) and .

**f) **The above works as a counterexample to this as well.

**10.**

**Problem: **Let be an infinite set and define by

a) Prove that this is a metric. b) Which subsets of the resulting metric space are open? c) Which are closed? d) Which are compact?

**Proof:**

**a) **Clearly and . Also, since equality and inequality are transitive symmetry immediately follows. The triangle inequality is just a set of three cases. If then . If then . If then .

**b) **For every we have that is open since . Thus, all subsets of being the union of singletons are open.

**c)** From b) it readily follows that every subset of is also open.

**d) ** will be compact if and only if is finite. To see this suppose that is infinite, then is evidently an open cover of it which admits no finite subcover. The converse is trivial.

**11.**

**Problem:** Let the following be maps from to , and determine if they determine a metric. a) , b) , c) , d) , e)

**Proof:**

**a)** This is not a metric.

**b) **This is a metric as one (I’m to lazy to tex it) can check.

**c) **Nope, .

**d) **Negatory, .

**e)** Yes. In fact, (once again I omit the proof) if is a metric so is

**12.**

**Problem:** Let with the usual topology be defined by . Prove directly (i.e. no Heine-Borel) that is compact.

**Proof: **This is true in any topological space . But, for the sake of convenience let us do it merely in a Hausdorff space . So, let is convergent sequence in with . Then, is a compact subspace of .

**Proof:** Let be an open cover for . Choose such that . Then, since all but finitely may values of lie in . So, for each of the other finitely many points of choose any element of which contains them. Clearly this will be a finite subcover.

Could you please make email contact with me? I’d like to pay you to tutor me via video chat. As you can tell from my email address, I am either a student at UA or a spambot disguised as a student! Thanks,

C

Comment by C W | October 7, 2010 |

I’m sorry friend, that doesn’t sound like a very good idea. I am way too busy this term.

Comment by drexel28 | October 7, 2010 |