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April 27, 2015

Posted by on 1. Show that a metric space is regular ( and normal ().

2. Show that a compact Hausdorff space is both regular and normal.

3. Consider with the lower limit topology. Which separation axioms (regular, normal, Hausdorff) does it satisfy?

4. Consider with the finite complement topology. Is it regular or normal?

5. Is with the lower limit topology second countable? Is it Lindelof?

6. What about with the finite complement topology? (same questions as 5)

7. A space is separable if it has a countable, dense subset. Is with the lower limit topology separable? What about the finite complement topology?

8. Show that all metric spaces are first countable, and that a separable metric space is second countable.

9. Show that if is a subset in a Hausdorff space, is a limit point of if and only if it is an accumulation point.

10. Show that “compact” implies “limit point compact”.

11. Show that limit point compactness plus Lindelof implies compactness.

12. Show that every open cover of a compact metric space has a Lebesgue number.

13. Show that the product of two compact spaces is compact.

14. Show that the product of a Linedolof space and a compact space is Lindelof. Note: the product of two Lindelof spaces need not be Lindelof.

15. Review the definition of locally compact. Show that is locally compact.

16-20 Show that the middle thirds Cantor set is

16. compact

17. totally disconnected

18 nowhere dense as a subset of (its closure has empty interior)

19. perfect (every point is an accumulation point

20. Homeomorphic to (review the proof)

21. Show that a finite set with the discrete topology is compact and that a countably infinite set with the discrete set is Lindelof.

22. Show that the rational numbers, in the subspace topology, is totally disconnected and dense (what is the closure?)

23. Show that if is the set of rational numbers, then is NOT compact (in the subspace topology).

24. Consider the real line (usual topology) with an extra point added: the system of open neighborhoods of this extra point consists of the complements of compact sets in the real line. Show that this new space is compact. What is it homeomrophic to?

**Definitions to know**

1. Everything prior to the first exam.

2. New:

Hausdorff, Regular (), Normal ( )

Compact

First Countable

Second Countable

Lindelof

Locally Compact (every point is lies in the interior of a compact set)

totally disconnected (components are one point sets)

0 dimensional (has a basis of clopen sets)

Nowhere dense subset (closure of the set has empty interior)

perfect (every point is a limit point)

accumulation point

condensation point (every open set contains an uncountable number of points of the given set)

Lebesgue number for an open cover in a metric space

Separable space (has a countable dense subset)

Cantor Set (topological characterization: compact, metric, totally disconnected, perfect)

Tychonoff Theorem (arbitrary product of compact sets is compact)

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April 27, 2015

Posted by on Here is the goal: if we have a compact metric space that is totally disconnected (components are one-point sets) and is perfect (every point of is an accumulation point of we want to show that is homeomorphic to a Cantor set.

Here is the plan:

1. Use the structure of to obtain a sequence of covers by clopen sets (sets both open and closed) which have diameters steadily shrinking to zero.

2. Use those covers to present as an inverse limit space.

3. Obtain a map to something called a derived sequences: this is basically an inverse limit space in which the spaces are discrete sets with the discrete topology.

4. Show that the map in step 3 is a homeomorphism.

5. Given a second space with the same property, repeat steps 1, 2, 3, except at step 3, we show that we can assume that the derived sequences have the same number of points.

6. Note that the two derived sequence spaces are homeomorphic because they have the same number of elements.

We need to start with a fact about compact metric spaces that are totally disconnected (Willard proves something more general here): if a compact metric space is totally disconnected, then it has a basis of clopen sets.

Note: a space which as a basis of clopen sets is called 0 dimensional.

Proof: let be an open set containing . Now consider . Since there exists disjoint open sets where these are the disjoint separating open sets that exist because is totally disconnected. Note that are clopen.

Because is a closed subset of a compact space, is compact so we need only a finite number of the to cover…say . Then is an clopen set (open by definition, closed because the complement is open) that contains and .

Ok, the preliminaries are set.

Step 1: let be our compact metric space which is totally disconnected and perfect. Let be another such space.

Cover by clopen sets whose diameters are less than 1. Because is compact, we can assume that only a finite number of clopen sets are used, say, . We now make this into a partition by clopen sets by setting so we have a partition by sets of diameter less than 1.

Now do the same for : obtain a partition by clopen sets . Now it is possible that . So, assume .

So, do the above procedure to split selected into disjoint clopen sets of smaller diameter and reindex the so that we have the same number of clopen sets in the first partition. That ends step 1. So have been partitioned into the same number of clopen sets of diameter less than 1.

Step 2: now do this again for diameter less than . But for start with the sets in and subpartition each into subsets of diameter less than to obtain a new subpartition (that is, gets sub partitioned, so does etc. Call these new sets where the . Do the same for , and, as in step 1, adjust so that the total number of elements of the second partition for is the same for .

Step 3: continue the process for all finite ; each clopen set in the partition has diameter less than lines in a unique set in the partition. The same holds for latex X_{\infty} $ described by and for . The maps are inclusion maps and are, in fact, homemorophisms.

Now we describe the “derived space”: remember each step partitions the space (or ) into sets of shrinking diameter. For each define a map that takes to a single point in the discrete space of elements. This map is clearly onto and continuous and maps to which is an inverse limit of discrete spaces; the maps in the discrete spaces are defined by inclusion; that is, if the maps are then where . That is, each point in the discrete space “to the right” is mapped uniquely to one in the left in a manner that makes the necessary diagram commute.

So the map between the inverse limit spaces is clearly a well defined, onto continuous map. Because is compact and is Hausdorff, we need only show that map is one to one. But that isn’t hard: if then for some and so lie in different partition sets in the level. So the map between and is a homeomorphism. But we get basically the same map between and because we chose the same number of partition elements for each .

Conclusion: all compact metric spaces that are both perfect and totally disconnected are homeomorphic. But the Cantor sets described in class are compact metrics spaces that are perfect and totally disconnected.

Therefore all compact metric spaces that are both perfect and totally disconnected are homeomorphic to Cantor sets. Hence it makes sense to talk about a Cantor Space.

April 22, 2015

Posted by on Reference: Chapter 8, sections 29-30 of *Willard*.

The ultimate goal is the following: we’d like to show that if is a compact metric space that is totally disconnected (components are one point sets) and is perfect (every point is an accumulation point) then is homeomorphic to (countable product of a two point set with the discrete topology); that is, is homemorphic to a Cantor set.

A valuable tool will be the concept of an “inverse limit space”, which we will now describe.

Let be topological spaces with, for all a continuous function. Now consider the following subset of This subspace (in the subspace topology) is called an *inverse limit space* and is determined by and is denoted .

The student should think of the following situation: remember how the Cantor middle third set was constructed? In this instance, think of as being , being and, in general, corresponding to . The maps are the “inclusion map” (the identity map restricted to .

This is a classic way of describing a space that is determined by a countable intersection of nested sets.

The elements of an inverse limit space are sometimes called *strings*.

Note: if one wants to build something as a countable union of sets, there is something called the *direct limit* that corresponds to that construction.

Back to our inverse limit space: the inverse limit space is sometimes empty. But it won’t be in our case for the following reason: if each in our inverse limit space is a non-empty compact, Hausdorff space, then our inverse limit space will also be a non-empty, compact Hausdorff space.

Here is how we show this: we show that this inverse limit space is the countable intersection of closed subsets of a compact Hausdorff space .

First, consider the sets . Note: these are points of the product space for which specifically at the k’th component; this relation doesn’t have to hold for the other components (though it can).

We claim that is a closed subset of . If then where . So find disjoint open sets in . Because is continuous, we can find an open set where and . Now which is an open set which misses

Now and each is a non empty closed subset of a compact Hausdorff space and the have the finite intersection property. So our is a nonempty, compact Hausdorff space.

**Wrapping your head around the concept**

To get a feel for what this is like, let us examine the case where is the Cantor set and with the maps being the inclusions.

Ask yourself the following questions:

1. What are the elements of ? Answer: if then . Example: is such an element. In fact ALL elements are where each .

2. What is an example of some element of that is NOT in ? Answer: here is one: but isn’t in . but is NOT in .

3. For question 2, find an element that in the complement of one of the and explicitly find an open set in which contains .

We aren’t quite ready to use these inverse limit spaces for what we want to do…just yet. So far, we’ve taken a situation and just put it into a different setting. First, we need to discuss continuous functions between inverse limit spaces. Then when we get to the derived sequence for certain spaces, we’ll be ready to make some headway. And the interesting thing is that our derived sequence will consist of a sequence of finite spaces with the discrete topology.

April 16, 2015

Posted by on This will look as if I’ve lost leave of my senses.

Start with a two point set with the discrete topology.

Now let where each is a two point set with the discrete topology (homeomorphic copy of ). Give the product topology.

1. Each is compact, hence it follows that is compact from the Tychonoff Theorem. However: prove that is compact by considering any open cover and showing that is open cover has a finite subcover. Hint: if , where each is open in . So, how many points are left uncovered by ?

2. This might seem like a silly exercise, but do it anyway: let Specifically provide two open sets, .

3. Show that is Hausdorff in the following way: if describe (mathematically) the disjoint open sets containing these two distinct points. Yes, I know that the product of Hausdorff spaces is Hausdorff; don’t use that result here.

4. Let be any bijection: where is a two point set with the discrete topology. Show that is a homeomorphism. Note: you’ve done more difficult problems than this one.

5. Now let by where is a bijection from the two point set to itself. Show that is a homeomorphism. Note: since we are talking about a bijection from a compact Hausdorff space to itself, we need only concern ourselves with be continuous.

6. This shows why no. 5 is important: given there exists a homeomorphism where .

Note: a topological space is called homogeneous if, given there exists a homeomorphism where .

Here is an example of a non-homogeneous space: consider in the subspace topology (usual topology).

7. Now here is the key result: show that and (the middle thirds Cantor set described here) are homeomorphic. Note: it isn’t that hard to get a set bijection between the two spaces; given one maps the binary sequence which uniquely describes to in the natural way. Since is compact and is Hausdorff, to show that the map is a homeomorphism, one needs only show that the given map is continuous. But that isn’t as hard as it might appear at first: just remember what an open set in the product topology consists of, and note what it “pulls back to” under your map. Also remember that at each stage, the disjoint intervals are clopen sets (both open and closed in the subspace topology).

In our next installment, we’ll give a couple of different manifestations of the Cantor set, which we can now call the “Cantor Space”.

April 15, 2015

Posted by on Read section 17.9, Chapter 6 of Willard.

Start with the interval . Call this

Let Give these two sets labels respectively.

Let . Give these sets labels:

respectively.

Note that removed the middle third from and removed the middle third from the two intervals of .

Inductively: let be with each middle third removed; note that consists of disjoint intervals, each of width . Note that where is a sequence of 0’s and 1’s which denote where the interval lies in the higher . Example: lies in .

Let .

Exercises:

1. Show that is closed, compact, and non-empty.

2. Show that is non empty.

3. Show that if then there is some where lie in different components of . (hint: if for some . Then….

4. Give the subspace topology (this is the usual one). Show that is totally disconnected; that is, for there exists open and .

5. Show that every is an accumulation point of . Show (in ).

6. Show that, as a subset of , has empty interior.

7. A subset is called *nowhere dense* if is empty. Show that as a subset of , is nowhere dense.

8. An interval is said to have measure . A set is said to have measure if where is any covering by intervals (closed, open, half open does not matter).

It is a fact that if with then (this requires proof). Assuming this fact, what is the measure of ? Hint: add up the measure of the “deleted intervals”: using provided .

9. Repeat the construction of but this time, delete the “middle .

That is, , etc.

Now do any of the answers for questions 1-7 change?

10. For the “middle 5’ths” Cantor set, what is the measure?

11. For any construct a Cantor “deleted interval” set of that measure.

We will show that the measure of a given Cantor set is NOT a topological property.

12. Show that these Cantor sets are uncountable. Hint: one way to do this is to use a result that we studied here. One other way is to exhibit a set bijection from your Cantor set to , which we’ve shown to be uncountable.

April 13, 2015

Posted by on We show that if is Lindelof and Hausdorff then limit point compactness implies compactness.

Assume that is Lindelof and Hausdorff and is limit point compact but not compact.

Let have an open cover which has no finite subcover; we can assume the cover to be countable by the Lindelof condition. So where each is open.

We assume that if then is not part of the cover; it is redundant. Now for each let . So is an infinite set, which has a limit point . We can assume that is also an accumulation point since is Hausdorff. Because the cover, for some , and then contains an infinite number of which must include for . But this is impossible since the were chosen to be disjoint from

Note: this is very similar to the argument in the David Wright paper on compactness in the section about “perfect limit points”.

April 13, 2015

Posted by on First of all, the remarks about the countability of can be seen here.

A topological space is said to be *second countable* if there is a countable basis. That is, there exists a countable collection such that, given any open set where .

Recall that a set is dense in if (that is, every point of is either in or a limit point of . If has a countable dense subset, then is said to be *separable*.

Note: , finite, in the usual topology is separable as the set , (the rational numbers) is both countable and dense.

**Theorem** If is metric and separable, then is second countable.

Proof: Let be the countable dense subset and then the set is a countable collection of open basis sets. Now if is any open set and then there exists some where . Now there is some where so .

It follows that if is separable but not second countable, there is no metric which produces the topology for ; that is, with that metric is not metrizable.

Example: , the real line with the lower limit topology, is not metrizable.

The proof: Note that is separable as is a countable, dense subset as every open set of the form contains a point of . But is NOT second countable. For if at least one of the . But there are an uncountable number of real numbers.

However does meet an countability condition of a sort. To see this, we’ll use a Lemma:

Lemma: a collection of disjoint open sets in a second countable space is countable. This is easy to see after a moment’s thought, as there can’t be more disjoint open sets than there are basis elements.

Now let be any open cover of . Let consist of all points not contained in the “usual topology interior of a . If is empty then is covered by the “interiors” of these open sets; hence a countable subcollection covers all of . Otherwise, note that is an open set in the “usual topology” which means that is closed. Now for all there is some where since any other does not lie in an interior. That means that these open intervals are disjoint and so there can only be a countable number of these. On the other hand is covered by open (in the usual topology) intervals and so a countable subcollection suffices for this subset.

Therefore any open cover of has a countable subcover. A space that has this property is called Lindelöf.

April 3, 2015

Posted by on I prefer the setting in Willard (19.2) because that description is more widely used.

So, consider with a given topology and an extra point . Now describe a topology for as follows: given let the open sets be which contain and let the open sets which contain be where is compact in . Call this topology which is called the *one point compactification topology*.

Theorem: is compact.

Proof. Let be any open cover for . At least one , say contains and then for some compact . So this covers all of except for . Now look at the rest of the open cover: covers which is compact and note each of these s is open in . So a finite number of these cover . So this finite subcollection together with is the required finite subcover.

Theorem: If is a locally compact, Hausdorff space, then is a compact Hausdorff space.

Proof. We need only prove the Hausdorff part. If we can separate them by disjoint open sets; we need only show that there are disjoint open sets separating from . Because is locally compact, there exists a compact set such that . Then and are the disjoint open sets containing and respectively.

Now we show that the one point compatification of is homeomorphic to .

First let be given by

We now produce two continuous maps which agree with each other where their domains join.

Let denote the part of at or below the hyperplane ; that is, that part of whose coordinate is negative:. Then is given by (this is the “straight up” projection.)

Note that so the image of is the closed “n-disk” .

Now map the upper part of to the complement of the unit n-disk in the hyperplane by:

Note that .

So both agree at the equator so the composite map is a continuous bijection.

Now extend to all of : by . has the one compactification topology. The extended is a continuous bijection.

But is compact and is Hausdorff (by the work at the start of this post) hence the extended is a homeomorphism.

April 3, 2015

Posted by on A topological space is called *perfect* if every is an accumulation point of . A subset is perfect if every point of is an accumulation point.

Examples: in the usual topology is perfect. (usual topology) is a perfect subset. is not a perfect subset. The rationals is a perfect subset.

We will show that a Cantor set is a perfect, compact set.

**Theorem** If is a perfect compact Hausdorff space, then is uncountable. If is a perfect compact subset of a Hausdorff space, then is uncountable.

Proof: we prove the first result and the second follows by using the subspace topology.

The plan: we let and show that there is always a point that is not in the image of . Denote by .

Consider . There exists an open . Because is a limit point, there exists some . Because is Hausdorff, there exists disjoint open sets separating from . Let and note that

Now consider : if then we can find disjoint open sets where contains and . So denote and note . Let and proceed with the next step.

If , note that is an open set containing which is a limit point. contains a second point of and we can obtain two disjoint open sets where and . Let and note .

Note that

This procedure can be repeated inductively for all and does not terminate. Now we note that is non empty because these are closed sets that have the finite intersection property and is compact and is disjoint from . Hence the image of any one to one map from the integers into cannot be onto; hence is uncountable.

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