Some questions

1. Show that a metric space is regular (T_3 and normal (T_4 ).

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

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

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

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

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

7. A space is separable if it has a countable, dense subset. Is R 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 C is a subset in a Hausdorff space, x is a limit point of C 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 R^n 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 [0,1] (its closure has empty interior)

19. perfect (every point is an accumulation point

20. Homeomorphic to 2^Z (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 Q is the set of rational numbers, then Q \cap [0,1] 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 (T_3 ), Normal (T_4 )
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)

Inverse Limit Spaces, derived sequences and equivalence of Cantor Sets

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

Here is the plan:

1. Use the structure of X 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 X 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 Y 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 U be an open set containing x . Now consider y \in Fr(U) . Since y \neq x there exists disjoint open sets U_y, V_y where (X \cap U_y) \cup (X \cap V_y) = X these are the disjoint separating open sets that exist because X is totally disconnected. Note that U_y, V_y are clopen.

Because Fr(U) is a closed subset of a compact space, Fr(U) is compact so we need only a finite number of the V_y to cover…say \cup_{i=1}^k V_{yk} . Then U'= U - \cup^k_{i=1} V_{yi} is an clopen set (open by definition, closed because the complement is open) that contains x and U' \subset U .

Ok, the preliminaries are set.

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

Cover X by clopen sets whose diameters are less than 1. Because X is compact, we can assume that only a finite number of clopen sets are used, say, V^1_1, V^1_2, ..V^1_{k_1} . We now make this into a partition by clopen sets by setting U^1_1 = V^1_1, U^1_2 = V^1_2 - V^1_1, ...U^1_i = V^1_i -(V^1_1 \cup V^1_2 \cup ...V^1_{i-1}) so we have a partition by sets of diameter less than 1.
Now do the same for Y: obtain a partition by clopen sets W^1_1, W^1_2, ...W^1_{j_1} . Now it is possible that j_1 \neq k_1 . So, assume j_1 > k_1 .

So, do the above procedure to split selected V^1_i into disjoint clopen sets of smaller diameter and reindex the V^1_i so that we have the same number of clopen sets in the first partition. That ends step 1. So X, Y 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 \frac{1}{2} . But for X start with the sets in V^1 and subpartition each V^1_i into subsets of diameter less than \frac{1}{2} to obtain a new subpartition (that is, V^1_1 gets sub partitioned, so does V^1_2, etc. Call these new sets V^2_{(i,j)} where the V^2_{(i,j)} \subset V^1_{i} . Do the same for W^1_k \subset Y , and, as in step 1, adjust so that the total number of elements of the second partition for X is the same for Y .

Step 3: continue the process for all finite k ; each clopen set in the k'th partition has diameter less than \frac{1}{k} lines in a unique set in the k-1'st partition. The same holds for X, Y.     Step 4:  Now form the inverse limit spaces latex X_{\infty} $ described by X_1 \leftarrow X_2 \leftarrow X_3 .... and for Y_{\infty} . The maps are inclusion maps and are, in fact, homemorophisms.

Now we describe the “derived space”: remember each step partitions the space X (or Y ) into sets of shrinking diameter. For each X_i = V^i_1 \cup V^i_2 \cup ...V^j_{j_i} define a map \phi_i that takes V^i_k to a single point in the discrete space of j_i elements. This map is clearly onto and continuous and maps X_{\infty} to Z_{\infty} which is an inverse limit of discrete spaces; the maps in the discrete spaces are defined by inclusion; that is, if the maps are h_i then h_i(\phi_i (V^i_j)) = h_{i-1}(\phi_{i-1}(V^{i-1}_p) where V^{i}_j \subset V^{i-1}_p . 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 X_{\infty} is compact and Z_{\infty} is Hausdorff, we need only show that map is one to one. But that isn’t hard: if x \neq y then d(x,y) > \frac{1}{m} for some m and so lie in different partition sets in the V^{2m} level. So the map between X_{\infty} and Z_{\infty} is a homeomorphism. But we get basically the same map between Y_{\infty} and Z_{\infty} because we chose the same number of partition elements for each m .

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.

Some illustrations of Cantor Sets in different contexts

cantor2

cantor3

cantor4

cantor5

cantor6

cantor7

cantor8

cantoreggs

cantorset

cantor9

cantor10

cantor11

Inverse limit spaces and their relevance to Cantor Spaces

Reference: Chapter 8, sections 29-30 of Willard.
The ultimate goal is the following: we’d like to show that if X is a compact metric space that is totally disconnected (components are one point sets) and is perfect (every point is an accumulation point) then X is homeomorphic to 2^Z (countable product of a two point set with the discrete topology); that is, X is homemorphic to a Cantor set.

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

inverselimit

Let X_0, X_1, X_2, ....X_n .... be topological spaces with, for all n \in \{ 1, 2, 3,... \}, f:X_{n} \rightarrow X_{n-1} a continuous function. Now consider the following subset of \Pi^{\infty}_{n=0} X_n : \{(x_0, x_1, x_2, x_3,....) | x_{n-1} = f(x_n), n \in \{1, 2, 3, ...\} \} This subspace (in the subspace topology) is called an inverse limit space and is determined by f, X_i and is denoted X_{\infty} .

The student should think of the following situation: remember how the Cantor middle third set was constructed? In this instance, think of X_0 as being C^0 , C^1 being X_1 and, in general, C^n corresponding to X_n . The maps are the “inclusion map” (the identity map restricted to C^n .

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 X_i 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 \Pi^{\infty}_{n=0} X_n .

First, consider the sets Y_k = \{x_0, x_1, x_2, ....x_{k-1}, x_k, x_{k+1},.... \} \in  \Pi^{\infty}_{n=0} X_n \} . Note: these are points of the product space for which x_{k-1} = f(x_k) specifically at the k’th component; this relation doesn’t have to hold for the other components (though it can).

We claim that Y_k is a closed subset of \Pi^{\infty}_{n=0} X_n . If z \notin Y_k then z = \{z_0, z_1,....z_{k-1}, z_k, ... \} where z_{k-1} \neq f(z_k) . So find disjoint open sets U, V in X_{k-1}, z_{k-1} \in U, f(z_k) \in V . Because f is continuous, we can find an open set W \subset X_{k} where z_{k} \in W and f(W) \subset V . Now z \in U \times W \times \Pi^{\infty}_{i \neq k-1, k} X_i which is an open set which misses Y_k

Now X_{\infty} = \cap^{\infty}_{i=1} Y_i and each Y_i is a non empty closed subset of a compact Hausdorff space and the Y_i have the finite intersection property. So our X_{\infty} 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 X is the Cantor set and X_k = C^k with the maps being the inclusions.

Ask yourself the following questions:

1. What are the elements of X_{\infty} ? Answer: if x \in X then x = (x, x, x, x, ....) . Example: (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}....) is such an element. In fact ALL elements are (\sum^{\infty}_{k=1}\frac{2x_k}{3^k}, \sum^{\infty}_{k=1}\frac{2x_k}{3^k}, \sum^{\infty}_{k=1}\frac{2x_k}{3^k}, ....) where each x_i \in \{0, 1 \} .

2. What is an example of some element of Y_k that is NOT in X_{\infty} ? Answer: here is one: (\frac{1}{3}, \frac{1}{3}, 0, 0, 0, ...) \in Y_1 but isn’t in X . (0, \frac{2}{3}, \frac{2}{3}, \frac{1}{3}, 0, \frac{1}{3},...) \in Y_2 but is NOT in X_{\infty} .

3. For question 2, find an element z \in \Pi^{\infty}_{k=0}X_k that in the complement of one of the Y_k and explicitly find an open set in \Pi^{\infty}_{k=0}X_k which contains z .

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.

Cantor Sets/Spaces Part II

This will look as if I’ve lost leave of my senses.
Start with T = \{0, 1 \} a two point set with the discrete topology.

Now let D = \Pi^{\infty}_{i=1} T_i where each T_i is a two point set with the discrete topology (homeomorphic copy of T ). Give D the product topology.

1. Each T_i is compact, hence it follows that D is compact from the Tychonoff Theorem. However: prove that D is compact by considering any open cover \mathscr{O} and showing that is open cover has a finite subcover. Hint: if O \in \mathscr{O} , O = U_1 \times U_2 \times... U_k \times \Pi^{\infty}_{i=k+1} T_i where each U_i is open in T_i . So, how many points are left uncovered by O ?

2. This might seem like a silly exercise, but do it anyway: let x= (1, 0, 0, 0, 0,....) \in D, y = (0, 0, 0, 0,...) \in D Specifically provide two open sets, U_x, x \in U_x, U_y, y \in U_y, U_x \cap U_y = \emptyset .

3. Show that D is Hausdorff in the following way: if x \neq y 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 f be any bijection: f:T \rightarrow T where T is a two point set with the discrete topology. Show that f is a homeomorphism. Note: you’ve done more difficult problems than this one.

5. Now let f:D \rightarrow D by f(x) = (f_1(x_1), f_2(x_2), ....,f_k(x_k),....) where f_k: T_k \rightarrow T_k is a bijection from the two point set to itself. Show that f is a homeomorphism. Note: since we are talking about a bijection from a compact Hausdorff space to itself, we need only concern ourselves with f be continuous.

6. This shows why no. 5 is important: given x, y \in D there exists a homeomorphism f:D \rightarrow D where f(x) = y .
Note: a topological space X is called homogeneous if, given x, y \in X there exists a homeomorphism f:X \rightarrow X where f(x) = y .

Here is an example of a non-homogeneous space: consider \{\frac{1}{n}, n\in \{1,2,3...\} \} \cup \{ 0 \} \} in the subspace topology (usual topology).

7. Now here is the key result: show that D and C (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 c \in C one maps the binary sequence which uniquely describes c to D in the natural way. Since C is compact and D 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 C^n 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”.

Cantor Set: Part I

Read section 17.9, Chapter 6 of Willard.

Start with the interval [0,1] . Call this C^0
Let C^1 = [0,\frac{1}{3}] \cup [\frac{2}{3}, 1] Give these two sets labels C^1_{(0)}, C^1_{(1)} respectively.

Let C^2 = [0, \frac{1}{9}] \cup [\frac{2}{9}, \frac{1}{3}] \cup [\frac{2}{3}, \frac{7}{9}] \cup [\frac{8}{9}, 1] . Give these sets labels:
C^2_{(0,0)}, C^2_{(0,1)}, C^2_{(1,0)}, C^2_{(1,1)} respectively.

Note that C^1 removed the middle third from C^0 and C^2 removed the middle third from the two intervals of C^1.

Inductively: let C^{n+1} be C^n with each middle third removed; note that C^n consists of 2^n disjoint intervals, each of width \frac{1}{3^n} . Note that C^n = \cup^n_{k=1} C^n_{\Pi^n_{\{x^k_i \}}} where \{x^k_i \} is a sequence of 0’s and 1’s which denote where the interval lies in the higher C^i . Example: C^4_{(1,0,1,0)} lies in C^1_{1}, C^2_{(1,0) }, C^3_{(1, 0, 1)} .

Let C = \cap^{\infty}_{k=1} C^k .

Exercises:

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

2. Show that C^1_{(x_1)} \cap C^2_{(x_1, x_2) } \cap C^3_{(x_1, x_2, x_3)} ...\cap C^k_{(x_1, x_2, x_3, ..., x_k)} \cap...... is non empty.

3. Show that if x, y \in C, x \neq y then there is some n where x, y lie in different components of C^n . (hint: if x \neq y, d(x,y) > \frac{1}{3^n} for some n . Then….

4. Give C the subspace topology (this is the usual one). Show that C is totally disconnected; that is, for x, y \in C, x \neq y there exists open U_x, U_y, x \in U_x, y \in U_y, U_x \cap U_y = \emptyset and C = (U_x \cap C) \cup (U_y \cap C) .

5. Show that every x \in C is an accumulation point of C . Show \overline{C} = C (in [0,1] ).

6. Show that, as a subset of [0,1] , C has empty interior.

7. A subset A \subset X is called nowhere dense if int(\overline(A)) is empty. Show that as a subset of [0,1] , C is nowhere dense.

8. An interval (a,b), b > a is said to have measure m((b-a))= b-a . A set A \subset R^1 is said to have measure m(A) if m(a) = lub(\sum_{j \in J} m(I_j) where \cup_{j \in J} I_j is any covering A by intervals (closed, open, half open does not matter).

It is a fact that if X \cup Y = B with X \cap Y = \emptyset then m(B) = m(X) + m(Y) (this requires proof). Assuming this fact, what is the measure of C ? Hint: add up the measure of the “deleted intervals”: \frac{1}{3} + 2 \frac{1}{9} + 4 \frac{1}{27} + 8 \frac{1}{81} .... using \sum^{\infty}_{k=0} r^k = \frac{1}{1-r} provided |r| < 1 .

9. Repeat the construction of C but this time, delete the “middle \frac{1}{5} .

That is, C^1 = [0, \frac{2}{5}] \cup [\frac{3}{5},1], C^2 = [0, \frac{1}{25}] \cup [\frac{9}{25}, \frac{2}{5}] \cup [\frac{3}{5}, \frac{16}{25}] \cup [\frac{24}{25}, 1] , 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 0 \leq m < 1 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 2^{Z} , which we’ve shown to be uncountable.

Limit point compactness in Lindelof spaces

We show that if X is Lindelof and Hausdorff then limit point compactness implies compactness.

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

Let X have an open cover which has no finite subcover; we can assume the cover to be countable by the Lindelof condition. So X = \cup^{\infty}_{k=1}U_k where each U_k is open.

We assume that if U_m \subset \cup^{m-1}_{k=1}U_k then U_m is not part of the cover; it is redundant. Now for each m let x_m \in U_m-(\cup^{m-1}_{k=1}U_k) . So \{x_m \} is an infinite set, which has a limit point x . We can assume that x is also an accumulation point since X is Hausdorff. Because the U_k cover, x \in U_n for some n, and then U_n contains an infinite number of x_i which must include x_i for i > n . But this is impossible since the x_i were chosen to be disjoint from \cup^{i-1}_{k=1} U_k

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

Separable, Second Countable and the Lindelöf property

First of all, the remarks about the countability of Z \times Z can be seen here.

A topological space X is said to be second countable if there is a countable basis. That is, there exists a countable collection \mathscr{B} such that, given any open set U, U = \cup^{\infty}_{i=1} B_i where B_i \in \mathscr{B} .

Recall that a set A \in X is dense in X if \bar{A} = X (that is, every point of X is either in A or a limit point of A . If X has a countable dense subset, then X is said to be separable.

Note: R^n , n finite, in the usual topology is separable as the set (q_1, q_2, ...q_n) \in R^n , q \in Q (the rational numbers) is both countable and dense.

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

Proof: Let Q \subset X be the countable dense subset and then the set B_q(\frac{1}{n}), n \in \{1, 2, 3....\}, q \in Q is a countable collection of open basis sets. Now if U is any open set and x \in U then there exists some n where B_x(\frac{1}{n}) \subset U . Now there is some q \in Q where q \in B_x(\frac{1}{10n}) so x \in B_q(\frac{1}{5n}) \subset B_x(\frac{1}{n}) .

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

Example: R^1_l , the real line with the lower limit topology, is not metrizable.

The proof: Note that R^1_l is separable as Q is a countable, dense subset as every open set of the form [a,b) contains a point of Q . But R^1_l is NOT second countable. For if [a, b) = \cup_{\beta \in I} [a_{\beta}, b_{\beta}) at least one of the a_{\beta} = a . But there are an uncountable number of real numbers.

However R^1_l 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 \mathscr{O} be any open cover of R^1_l . Let F \subset R^1 consist of all points not contained in the “usual topology interior of a [a,b) \in \mathscr{O} . If F is empty then R^1 is covered by the “interiors” of these open sets; hence a countable subcollection covers all of R^1 . Otherwise, note that R^1 - F is an open set in the “usual topology” which means that F is closed. Now for all x \in F there is some \delta > 0 where (x, x+\delta) \cap F = \emptyset since any other y \in F 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 R^1 - F is covered by open (in the usual topology) intervals and so a countable subcollection suffices for this subset.

Therefore any open cover of R^1_l has a countable subcover. A space that has this property is called Lindelöf.

One point compactification topology

I prefer the setting in Willard (19.2) because that description is more widely used.
So, consider X with a given topology \mathscr{T} and an extra point \{p \} . Now describe a topology for X \cup \{p \} as follows: given x \in X let the open sets be U \in \mathscr{T} which contain x and let the open sets which contain p be X - C where C \subset X is compact in (X, \mathscr{T} ) . Call this topology \mathscr{T}_c which is called the one point compactification topology.

Theorem: X \cup \{p \} is compact.

Proof. Let \mathscr{O} be any open cover for X \cup \{ p \} . At least one O \in \mathscr{O} , say O_p contains p and then O_p = X - C for some compact C \subset X . So this O_p covers all of X except for C . Now look at the rest of the open cover: U_{O\in \mathscr{O}, C \cap O \neq \emptyset}O covers C which is compact and note each of these O s is open in X . So a finite number of these cover C . So this finite subcollection together with O_p is the required finite subcover.

Theorem: If X is a locally compact, Hausdorff space, then X \cup \{p \} is a compact Hausdorff space.

Proof. We need only prove the Hausdorff part. If x \neq y, x, y \in X we can separate them by disjoint open sets; we need only show that there are disjoint open sets separating x \in X from p . Because X is locally compact, there exists a compact set C such that x \in int(C) . Then X - C and int(C) are the disjoint open sets containing p and x respectively.

Now we show that the one point compatification of R^n is homeomorphic to S^n .
First let S^n \subset R^{n+1} be given by \{ (x_1, x_2, ...x_n, x_{n+1}) | (x_1)^2 + (x_2)^2 +...+(x_n)^2 + (x_{n+1})^2 = 1 ) \}

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

Let S^- denote the part of S^n at or below the hyperplane (x_1, x_2, ....x_n, 0 ) ; that is, that part of S^n whose x_{n+1} coordinate is negative:. Then f:S^- \rightarrow R^n is given by f((x_1, x_2, ....x_n, x_{n+1})) = (x_1, x_2, x_3, ....x_n, 0 ) (this is the “straight up” projection.)

Note that (x_1)^2 + (x_2)^2 +....(x_n)^2 = 1 - (x_{n+1})^2 so the image of f is the closed “n-disk” (x_1)^2 + (x_2)^2 +....(x_n)^2 \leq 1 .

Now map the upper part of S^n - \{(0, 0,....,0, 1) \} to the complement of the unit n-disk in the x_{n+1} = 0 hyperplane by:

upperspehere

g(x_1, x_2, ...x_n, x_{n+1}) = (\frac{x_1}{1-x_{n+1}}, \frac{x_2}{1-x_{n+1}}, ....\frac{x_n}{1-x_{n+1}}, 0) Note that 0 \leq x_{n+1} < 1 .

So both f, g agree at the equator so the composite map \phi: (S^n - \{(0, 0, 0,...0, 1) \} ) \rightarrow R^n is a continuous bijection.
Now extend \phi to all of S^n : \phi: S^n \rightarrow R^n \cup \{p \} by \phi ((0,0,0...0,1)) = \{p \} . R^n \cup \{p \} has the one compactification topology. The extended \phi is a continuous bijection.

But S^n is compact and R^n \cup \{ p \} is Hausdorff (by the work at the start of this post) hence the extended \phi is a homeomorphism.

Special result: perfect compact spaces and sets

A topological space X is called perfect if every x\in X is an accumulation point of X . A subset C \subset X is perfect if every point of C is an accumulation point.

Examples: R^n in the usual topology is perfect. B_x(\epsilon) \subset R^n (usual topology) is a perfect subset. \{ \frac{1}{n} \}, n \in \{1,2,3... \} \} is not a perfect subset. The rationals Q \subset R^1 is a perfect subset.

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

Theorem If X is a perfect compact Hausdorff space, then |X| is uncountable. If C is a perfect compact subset of a Hausdorff space, then |C| is uncountable.

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

The plan: we let f: Z \rightarrow X and show that there is always a point x \in X that is not in the image of f . Denote f(n) by x_n .

Consider x_1 . There exists an open U, x_1 \in U . Because x_1 is a limit point, there exists some y \neq x, y \in U . Because X is Hausdorff, there exists disjoint open sets U_1, W_1 separating x_1 from y_1 . Let V_1 = W_1 \cap U and note that x_1 \notin \overline{V_1}

Now consider x_2 : if x_2 \neq y_1 then we can find disjoint open sets U_2, W_2 where U_2 contains x_2 and y_1 \in W_2 \subset V_1 . So denote W_2 = V_2 and note \overline{V_2} \subset \overline{V_1} . Let y_2 = y_1 and proceed with the next step.

If x_2 = y_1 , note that V_1 is an open set containing y_1 which is a limit point. V_1 contains a second point of y_2 and we can obtain two disjoint open sets U_2, W_2 where y_2 \in W_2 and x_2 \notin \overline{W_2} . Let V_2 = W_2 and note \overline{V_2} \subset \overline{V_1} .

Note that \{ x_1, x_2 \} \cap (\overline{V_1} \cap \overline{V_2}) = \emptyset

This procedure can be repeated inductively for all k \in \{1,2,3,... \} and does not terminate. Now we note that \cap^{\infty}_{k=1} \overline{V_k} is non empty because these are closed sets that have the finite intersection property and X is compact and f(Z) = \cup^{\infty}_{i=1} x_i is disjoint from \cap^{\infty}_{k=1} \overline{V_k} . Hence the image of any one to one map from the integers into X cannot be onto; hence X is uncountable.