Bradley University Topology Class

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.

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.

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.

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:

$g(x_1, x_2, ...x_n, x_{n+1}) = (\frac{x_1}{1-x_{n+1}}, \frac{x_2}{1-x_{n+1}}, ....\frac{x_n}{1-x_{n+1}}, 0)$ 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.