An Introduction to Analysis/Sets and topology

Sets

It is tricky to define sets. For one thing, throughout the book, we define every mathematical object (e.g., function) in terms of sets or those who we defined with sets, and hence the problem: how to initiate the string of definitions. Figuring out how this can be done is a question in the set theory proper or logic and those subjects will be not be covered in the book. Lucky for us, we can still study most of problems in analysis with native understanding of sets, except for Axiom of Choice, which we shall prove must hold for certain indispensable results to be true.

The notion of "classes" is essential in formulating category theory. We define a class with the following properties:

• (i) Classes with the same elements are the same.
• (ii) A class $x$ is a set $y$ if there is a class $y$ with $x\in y$; otherwise, a class is said to be proper
• (iii) There is a class $\{x;\phi (x)\}$ where $\phi$ is a formula over sets $x$.

See also w:glossary of category theory for the terminology.

Here, a set is understood in a usual axiomatic set theory. For example, a class $\{x;x\in ̸x\}$ is proper.

We say a nonempty collection $ℱ$ of subsets of a set $S$ is a filter over $S$ if

• (1) $0\in ̸ℱ$
• (2) If $A$ and $B$ are in $ℱ$, then $A\cap B\in ℱ$.
• (3) If $A\in ℱ$ and $A\subset B\subset S$, then $B\in ℱ$.

The collection $ℱ$ is called an ultrafilter if, in addition to the above,

• (4) If $A\cup B\in ℱ$, then either $A\in ℱ$ or $B\in ℱ$.

In particular, that every filter is nonempty and satisfying (3) implies that $0^c$ is in each filter. We basically view that a filter is a generalized version of neighborhoods of a point (i.e., open sets containing the point). In fact, given a point $x$ in some set $S$, let $ℱ=\{A\subset S:x\in A\}$. Then $ℱ$ is an ultrafilter on $S$. For the similar reason the intersection of filters is again a filter.

We say that a collection $𝒮$ has the finite intersection property if $E\subset S$ is finite, then $E$ has nonempty intersection.

1.1 Lemma Every nonempty subset of a filter has the finite intersection property.
Proof: Let $ℱ$ be a filter and $E\subset ℱ$ be nonempty. If $D\subset E$ is nonempty and finite, then $\cap D$ is in $ℱ$ and is thus nonempty. Hence, $E$ has the finite intersection property. $◻$

1.2 Theorem Let $ℱ$ be a nonempty collection. Then the following are equivalent:

• (i) $ℱ$ is an ultrafilter.
• (ii) If $ℱ\subset 𝒢$ and $ℱ$ and $𝒢$ have the finite intersection property, then $ℱ=𝒢$. (i.e., $𝒢$ is maximal.)
• (iii) $ℱ$ is a filter and if $A\cup B\in ℱ$, then either $A\in ℱ$ or $B\in ℱ$.

Proof: Suppose (i). If $A\subset 𝒢$, then since $ℱ$ is an ultrafilter, $A$ or $A^c$ is in $ℱ$. If $A^c\in F$, then that means $A\cap A^c=0\in ℱ$, contradicting that $F$ has the finite intersection property. Thus, (i) $\Rightarrow$ (ii). To show (ii) implies (iii), let $𝒢$ consist of sets $A$ such that $A_1\cap A_2...\cap A_n\subset A$ for some nonempty finite sequence $A_1,A_2,...A_n\in ℱ$. Claim: $𝒢$ is a filter. Indeed, (1) $0\in ̸𝒢$ since no member of $ℱ$ is the empty set. (2) For $A,B\in ℱ$ both $A$ and $B$ must contain the intersection of the two intersections of finite subsets of \mathcal{G}</math>; thus, $A\cap B\in ℱ$. (3) For $A\in ℱ$ and $A\subset B$ since $A$ contains the finite intersection so does $B$; hence, $B\in ℱ$. Using (ii) we have: $ℱ=𝒢$. Let ${ℱ}_A=ℱ\cup \{A\}$ and ${ℱ}_B=ℱ\cup B$ and claim: either ${ℱ}_A$ or ${ℱ}_B$ has the finite intersection property. To get a contradiction, suppose not; that is, we can find $C$ and $D$ in $ℱ$ such that $C\cap A=0=D\cap B$. But this then means: by the distributive law $(C\cap D)\cap (A\cup B)=C\cap ((D\cap A)\cup (D\cap B))=C\cap D\cap A=$ empty, a contradiction to the axioms of filters since both $C\cap D$ and $A\cup B$ are in $ℱ$. Since the claim holds, appealing to (ii) again we get either $ℱ=ℱ\cup \{A\}$ or $ℱ=ℱ\cup \{B\}$. We conclude that (ii) implies that (iii). Finally, (iii) implies (i) since for any $A$ $A\cup A^c=0^c\in ℱ$ and by (iii) $A$ or $A^c$ is in $ℱ$. $◻$

With regard to (ii) in the theorem, the real question is if such an ultrafilter indeed exists, which we shall show cannot be unprovable without Axiom of Choice.

1 Lemma If $ℱ$ is a collection and has nonempty intersection, then $ℱ$ can be extended to an ultrafilter.
Proof: By assumption $\cap ℱ$ has a point $x$. Then a ultrafilter generated at $x$ is an ultrafilter containing $\cap ℱ$. $◻$

We assume the Axiom of Choice which states this. Let $𝒩$ be a nonempty infinite collection of pairs such that if $A,B\in 𝒩$, then $A$ and $B$ are disjoint. Then there exists a set consisting of exactly one element from each set in $𝒩$.

It is difficult to give some "constructive" example of a case for it is meaningful to apply AC. For example (the example is due to Bertrand Russell), let $𝒩$ be a collection of pairs of shoes. Then we can simply let S be the set of left shoes; this can be done without AC. The chief use of AC is, in other words, to state an existence of a some set when any constructive approach like the above example is impossible. Since AC is merely an assumption, after invoking AC, we would have no idea how choice is done; that is, how each element is chosen from a set and the set that resulted is never unique. If we apply AC to a collection of pairs of shoes, we would not know which shoe would be chosen. This is the chief reason for the criticism of AC since there seems no intuitive or constructive supports to assume AC. I only go as far to say that the problem has more to do with philosophy and logic and today most mathematicians are comfortable with assuming AC. See, for instance, [1]

1 Theorem The following are equivalent:

• (i) Every boolean algebra has a prime ideal. (Boolean Prime Ideal Theorem)
• (ii) Every ideal on a boolean algebra B can be extended to a prime ideal. (Strong Boolean Prime Ideal Theorem)
• (iii) Every filter over a set S can be extended to an ulterfilter over S. (Ultrafilter lemma)

Proof: Exercise. $◻$

If $ℱ$ is a filter and $x\in E\Rightarrow E\in ℱ$, then we say $ℱ$ converges to $x$ (not necessarily uniquely).

Topological space

For an arbitrary set $s$ a topology is a subset of a power set of $s$ that includes the empty set, $s$, the union of any subset of $\tau$ and the intersection of any two members of $\tau$. The members of $\tau$ are said to be open in $s$.

A set $E$ is closed in $G$ when its complement in $G$; i.e. $G/E$ is open. The empty set is denoted by $\varnothing$. Both $G$ and $\varnothing$ are open in $G$ since both are in $\tau$, and are also closed since $G\mathrm{\setminus }G=\varnothing$ and $G\mathrm{\setminus }\varnothing =G$, both of which are open.

In usual cases, we are given some $G$ and then induce a topology to $G$ by defining a $\tau$ so as to meet our needs at hand. We thus call $\tau$ a topology for $G$. The key insight here is that saying some set is open and another closed is merely the matter of labeling. By inducing a topology, we label some sets to be open, then the rest are deemed closed. Indeed, for example, by taking complement of the topology, one can always induce it as another topology where an open set becomes closed and a closed set open.

To give some concrete example, consider the set $1,2,3$ then we can induce a topology by defining, for example, $\tau =\{\},\{1\},\{1,2,3\}$. Can you induce another topology to the set? If so, how many possible topology for this set are there? The example shows that topologies for the same set can be larger or smaller. To give formal definition, suppose ${\tau }_1$ and ${\tau }_2$ are topologies. If ${\tau }_1\subset {\tau }_2$, then we say ${\tau }_1$ is said to be coarser than ${\tau }_2$ and ${\tau }_2$ is finer than ${\tau }_1$. For every set $G$, the finest topology that can be induced is the collection of all subsets of $G$ while the coarsest is the collection of the empty set and $G$. Both the topologies are of little interest.

The closure $\bar{E}$ of $E$ in $G$ is the intersection of all closed sets in $G$ containing $E$. The interior of $E$ is the union of all open sets contained in $E$. Clearly, interior of $E\subset E\subset \bar{E}$. The boundary $\text{b}E$ of $E=\bar{E}\mathrm{\setminus }\text{int}E$. It follows that a set has empty boundary if and only if it is open and closed. We say $E$ is dense in $G$ if $\bar{E}=G$ where the closure is taken in $G$. Equivalently, a set $E$ is dense in $G$ if $\text{b}E=G\mathrm{\setminus }E$.

1 Lemma (i) $\cup \overline{E_{\alpha }}\subset \overline{\cup E_{\alpha }}$ where the inequality holds if the index set of $\alpha$ is finite and (ii) $\overline{\cap E_{\alpha }}\subset \cap \overline{E_{\alpha }}$.
Proof: (i) Since $\overline{E_{\alpha }}\subset \overline{\cup E_{\beta }}$ for all $\alpha$ the desired inequality holds, and since the finite union of closed sets is closed the equality holds for a finite index set. The similar argument shows (ii).$◻$

1 Lemma The intersection of topologies for the same set is again a topology for the set.
Proof: Let $ℱ$ be a family of topologies for the same set. If $\tau \subset \cap ℱ$, then $\tau \subset$ every member of $ℱ$, which is closed under unions. Thus, $\cup \tau$ is in every member of $ℱ$. The other properties can be verified in the same manner. $◻$

Given a set $E$, the lemma says that the intersection of the collection of all topologies containing $E$ is a topology, which is the weakest by definition. We say this topology is generated by $E$.

We say a point $p$ is a limit point of a set $E$ if every neighborhood of $p$ contains $q\ne p$ such that $q\in E$.

1 Theorem A set $E$ is closed if and only if it contains all of its limit points.
Proof: The following are equivalent.

• (i) Every point in $E^c$ is not a limit point of $E$.
• (ii) Every point in $E^c$ has a neighborhood disjoint from $E$.
• (iii) $E^c$ is open.

$◻$

Functions

A function $f$ is a nonempty set of pairs such that $(a,b),(a,c)\in f$ implies that $b=c$ (i.e., single-valued). We then write $f(a)=b$ for a pair $(a,b)\in f$. While for a function, say $f(x)=2x$, we know how it maps elements; i.e., 1 goes to 2, $\pi$ goes to $2\pi$, etc, there need not be a way to specify precisely (i.e., closed form) how a function maps each element. For example, consider a set of infinitely many pairs (1, random), (2, random), .... This is a function by definition, but it takes infinitely many statements to state this particular definition of the function since randomness.

The domain of a function $f$ is the set of all first elements of pairs in $f$. For a set $A$, we define $f(A)=\{f(x):x\in A\}$, regrettably though this syntax is ambiguous on some occasions. We write $f:A\to B$ to say that $A$ is the domain of $f$ and $f(a)\in B$ for all $a\in A$. The image of a function $f$ then is the set $f(A)$, which is necessarily a subset of $B$. Clearly, the image of $f$ is a subset of $B$ but not need to be the same as $B$. When this equality, however, holds, we say that $f$ is surjective or a surjection. Let $f:A\to B$. Then the pre-image of $C$ under $f$, denoted by $f^{-1}(C)$, is the set of all $x\in A$ such that $f(x)\in C$. A restriction of $f:A\to B$ to $C\subset A$, denoted by $f{\mid }_C$, is a function $g:C\to B$ such that $g(x)=f(x)$ for every $x\in C$. Likewise, an extension of $f:A\to B$ to $A\subset C$ is any function $g:C\to B$ such that $g(x)=f(x)$ for every $x\in A$. Thus, every restriction is necessarily unique, while an extension might not be in general.

For example, let $g(1)=10,g(2)=20,g(3)=20$. Then we have:

$g=\{(1,10),(2,20),(3,20)\}$,

and the domain (resp. the image) of $g$ is $\{1,2,3\}$ (resp. $\{10,20\}$). The function $g$ is an example of a non-injection, while the restriction $h$ of $g$ to $\{1,2\}$ is injective.

1.1 Theorem Let $\pi$ be a function. The following hold:

• (i) $(g\circ f^{-1})\circ h=g\circ (f^{-1}\circ h)$
• (ii) $A\subset f^{-1}(f(A))$
• (iii) $f(f^{-1}(A))\subset A$
• (iv) $\bigcup _{a\in A}{f}^{-1}(E_{\alpha })=f^{-1}(\bigcup _{a\in A}{E}_{\alpha })$
• (v) $f^{-1}(A^c)=(f^{-1}(A){)}^c$

Proof: (i) Obvious from the definition. (ii) if $x\in f^{-1}(f(A))$, then $f(x)\in f(A)$ and thus $x\in A$. (iii) Let $y\in f(f^{-1}(A))$. Then we can find a $x\in f^{-1}(A)$ so that $y=f(x)$. It then follows that $y=f(x)\in A$. (iv) and (v) are obvious from definition. $◻$

We say $f$ is injective if for every $x,y\in A$ $f(x)=f(y)$ implies $x=y$. The definition is equivalent to say $f^{-1}$ is single-valued on the image $C$ of $f$(i.e., a function) since for $x_1,x_2\in f^{-1}(C)$, $f(x_1)=f(x_2)$ implies that $x_1=x_2$. $f$ is bijective if it is both injective and surjective. We write $f\circ g(x)=f(g(x))$ and this is called a composition. Compositions need not commute; for instance, let $f(x)=3$ and $g(x)=5$. Then $g\circ f=5\ne 3=f\circ g$. The identity function, or the identity for short, ${\text{id}}_A:A\to A$ is defined by $\text{id}(x)=x$ for all $A\in x$.

1.2 Theorem Suppose we have $f:A\to B$. Then the following hold:

• (i) f is injective if and only if there exists a function $g:B\to A$ such that $g(f(x))=x$ for every $x\in A$.
• (ii) f is surjective if and only if there exists a function $g:B\to A$ such that $f(g(x))=x$ for every $x\in B$. (assuming Axiom of Choice)
• (iii) f is bijective if and only if there exists a function $g:B\to A$ such that $g\circ f=i{d}_B$ and $f\circ g=i{d}_A$.

Proof: (i) The converse first; suppose $g\circ f=i{d}_A$. If $f(x_1)=f(x_2)$, then taking $g$ on both sides we get: $g\circ f(x_1)=g\circ f(x_2)$. Since $g\circ f$ is the identity on $A$, $x_1=x_2$. Hence, $f$ is injective. Conversely, suppose $f$ is injective and $C$ is the image of $f$. We may suppose that $A$ is non-empty and thus we find some $z\in B$. Since $f^{-1}$ is a function on $C$, let $g(y)=f^{-1}(y)$ if $y\in C$ and $g(y)=z$. Then $g\circ f$ is the identity. (ii) Suppose $f$ is surjective; thus, $f^{-1}(\{y\})$ has at least one element for every $y\in B$. Invoking Axiom of Choice, let $g(y)$ be one particular $x\in A$ "chosen" from $f^{-1}(\{y\})$. Conversely, suppose $f$ is not surjective, then clearly the image of $f$ restricted to any set is not $B$. (iii) using (i) and (ii) we find $g,h:B\to A$ so that $g\circ f={\text{id}}_A$ and $f\circ h={\text{id}}_B$. But $g=g\circ f\circ h=h$. $◻$

Finally we have:

1 Axiom A class $c$ is proper if and only if there is a bijection between $s$ and the universe.

1 Theorem Let $A,B$ be sets. If there are injections from $A$ to $B$ and from $B$ and $A$, respectively, then there is a bijection between $A$ and $B$
Proof (from [2]):

Continuity

Let $f:A\to B$. We say $f$ is continuous when its pre-image of every open set in $B$ is open, or equivalently, $\overline{f^{-1}(\stackrel{‾}{E})}=f^{-1}(\bar{E})$ for every $E\subset B$.

1.4 Theorem Let $f:A\to B$. The following are equivalent:

• (i) $f$ is continuous.
• (ii) $f(\bar{E})\subset \overline{f(E)}$ for every $E\subset A$.
• (iii) Let $x\in A$. If $f(x)\in G$ and $G$ is open in $B$, then there exists a $N\subset A$ open such that $x\in N$ and $f(N)\subset G$.

Proof: Suppose (i). Since $\overline{f^{-1}(F)}=f^{-1}(F)$ for $F\subset B$ closed, we have:

$\bar{E}\subset \overline{f^{-1}\circ f(E)}\subset \overline{f^{-1}(\stackrel{‾}{f(E)})}=f^{-1}(\overline{f(E)})$.

Then taking $f$ on both sides shows that (i) $\Rightarrow$ (ii). Suppose (ii). $◻$

1.4 Theorem The composite of continuous functions is again continuous.
Proof: Let $f:A\to B$ and $g:B\to C$ and suppose $f$ and $g$ are both continuous. Let $G\subset C$ be open. Since $g$ is continuous, $g^{-1}(G)$ is open in $B$. Since $f$ is continuous, $f^{-1}\circ g^{-1}(G)$ is open in $A$. It hence follows that $g\circ f$ is continuous on $A$ since $(g\circ f{)}^{-1}$ is an open mapping. $◻$

1 Theorem Let $f,g:\bar{A}\to B$ be continuous on $A$. If $f=g$ on $A$, then $f=g$ on $\bar{A}$.
Proof:

1.5 Theorem Every continuous function maps a compact set to a compact set.
Proof: Let $K$ be compact and $\{G_{\alpha }\}$ be an open cover of $f(K)$. We thus have: upon taking $f^{-1}$ on both sides:

$(f^{-1}\circ f)(K)=K\subset \bigcup _{\alpha }{f}^{-1}(G_{\alpha })$.

Here each $f^{-1}(G_{\alpha })$ is open since the continuity of $f$. Taking $f$ on both sides then gives the theorem. $◻$

Let $f$ be a function to some topological space. Then the set of $f^{-1}(G)$ for every open set $G$ is a the weakest among topologies that make $f$ continuous. If $ℱ$ is a family of functions defined on the same set, then a weak topology generated by $ℱ$ is the intersection of the weakest topology that makes each member of $ℱ$ continuous. A weak topology is indeed a topology since the intersection of topologies for the same set is again a topology by Lemma.

Compact sets

A cover of the set E is a collection of sets {$G_{\alpha }$} such that $E\subset \bigcup G_{\alpha }$. An open cover of $E$ is a cover of $E$ consisting of open sets, or equivalently, a subset of the topology whose union contains $E$. A subset of cover of $E$ is called a subcover if it is again a cover of $E$.

A set $K$ is compact if every open cover {$G_{\alpha }$} of it has a finite subcover; i.e.,

$K\subset G_1\cup G_2\cdots \cup G_n\subset \bigcup G_{\alpha }$ for some $n$.

For example, let $\{G_{\alpha }\}$ be open cover of the finite set $\{a_1,a_2,...a_n\}$. Then for each $a_k$, we can find some $G(a_k)$. It thus follows that a finite set (e.g., the empty set) is compact since

$\{a_1,a_2,...a_n\}\subset G(a_1)\cup G(a_2)\cup ...G(a_n)\subset \bigcup G_{\alpha }$

1.3 Theorem Let $K$ be compact. All of the following sets are compact:

• (a) A closed subset $F$ of $K$.
• (b) The union of $K$ and a compact set $L$.

Proof: (a) Let $\{G_{\alpha }\}$ be an open cover of $F$. Then since $F^c$ is open, we have:

$K\subset F^c\cup \bigcup G_{\alpha }$.

Since $K$ is compact, we find a finite subcover and we get

$F\subset K\subset F^c\cup G_1\cup ...G_n$.

Taking the intersection with $F$ shows that $F$ is compact. (b) Let $\{G_{\alpha }\}$ be an open cover of $K\cup L$. Then since $\{G_{\alpha }\}$ is an open cover of both $K$ and $L$, we find two finite subcover of $K$ and $L$, respectively. It then follows:

$K\cup L\subset G(K{)}_1\cup ...G(K{)}_n\cup G(L{)}_1\cup ...G(L{)}_m$. $◻$

1 Theorem $K$ is compact if and only if every collection of closed subsets of $K$ over which any finite intersection is nonempty has nonempty intersection.
Proof: Consider the contrapositive of the second part:

Every collection of closed subsets of $K$ with empty intersection admits a finite sub-collection with empty intersection.

If $\mathrm{\Gamma }$ is an open cover of $K$, then the collection $\{G^c\cap K;G\in \mathrm{\Gamma }\}$ is a collection as per the above statement; hence, $\mathrm{\Gamma }$ admits a finite subcover if and only if the statement is true. $◻$

1 Corollary If $K_n\supset K_{n_1}\supset ...$ is a sequence of compact sets, then $\underset{n=1}{\overset{\mathrm{\infty }}{\cap }}{K}_n$ is nonempty.

1 Theorem If a topological space is the union of countably many compact sets, then any of its open cover admits a countable subcover.
Proof: Let $K_j$ be a sequence of compact sets, and suppose that its union is covered by some open cover $\mathrm{\Gamma }$. For each $j=1,2,...$, since $\mathrm{\Gamma }$ is an open cover of $K_j$, it admits a finite subcover ${\gamma }_j$ of $K_j$. Now, ${\gamma }_1\cup {\gamma }_2...$ is a countable subcover of $\mathrm{\Gamma }$. $◻$.

1 Theorem The following are equivalent:

• (i) Axiom of Choice.
• (ii) Every product topology of compact topologies is compact. (Tychonoff's product theorem)
• (iii) Something has empty intersection.

Proof: Let $X$ be a product topology and $\{{\pi }_{\alpha }\}$ be a collection of projections on $X$. Let $ℱ$ be an ultrafilter on $X$. For each $\alpha$, since ${\pi }_{\alpha }(ℱ)$ is again an ultrafilter and ${\pi }_{\alpha }(X)$ is compact, ${\pi }_{\alpha }(ℱ)$ converges. From the lemma 1.something it follows that $ℱ$ converges. If (i) is true, then the convergence implies that $X$ is compact. To show (ii) implies (iii), Let $\{X_{\alpha }\}$ be a nonempty collection of nonempty sets. Also, let $a$ be a element such that $a\in ̸\cap X_{\alpha }$. Such an element must exist since the contrary means that $\cap X_{\alpha }$ is the universal set. For each $\alpha$ let $Y_{\alpha }=X_{\alpha }\cup \{a\}$ and induce the topology ${\tau }_{\alpha }$ by letting ${\tau }_{\alpha }=\{0,\{a\},X_{\alpha },Y_{\alpha }\}$. Then since its topology is finite, each $Y_{\alpha }$ is compact. That (iii) implies (i) is well known in set theory. $◻$

1 Theorem (Tychonoff's product theorem) The following are equivanelt:

• (i) Every product space of compact spaces is compact.
• (ii) Axiom of Choice.

Proof: (i) $\Rightarrow$ (ii). Let $\{X_{\alpha }{\}}_{\alpha \in A}$ be a collection of compact spaces and ${\pi }_{\alpha }$ be a projection from $\prod X_{\alpha }\to X_{\alpha }$. Let $ℱ$ be an ultrafilter on $\prod X_{\alpha }$. For each $\alpha$, since ${\pi }_{\alpha }(ℱ)$ is again an ultrafilter and $X_{\alpha }$ is compact, ${\pi }_{\alpha }(ℱ)$ must converge. Then it follows that $ℱ$ converges. Axiom of Choice implies that the product space is compact. Conversely, let $\{X_{\alpha }{\}}_{\alpha \in A}$ be a nonempty collection of nonempty sets. Let $p$ be a point such that $p\in X_{\alpha }$ for all $\alpha$. Such a $p$ must exist; if not, the intersection of $X_{\alpha }$ is the universal set, contradicting that it is a proper class. For each $\alpha$, let $Y_{\alpha }=X_{\alpha }\cup \{p\}$ and ${\tau }_{\alpha }=\{0,p,X_{\alpha },Y_{\alpha }\}$. Then ${\tau }_{\alpha }$ is a topology for $Y_{\alpha }$ and since finiteness is compact. Using (i) $\prod Y_{\alpha }$ is compact and thus $\{X_{\alpha }\}$ has the finite intersection property and this implies the statement equivalent to (ii). $◻$

We say a point is isolated if the set $x$ is both open and closed, otherwise called an limit point. If a set has no limit point, then the set is said to be discrete.

1 Lemma Every infinite subset $E$ of a compact space $K$ has a limit point.
Proof: Let $E\subset K$ be infinite and discrete. Then $E$ is closed since it contains all of limit points of $E$. Since $E$ is a closed subset of a compact set, $E$ is compact. It now follows: for each $x\in E$, the singleton $\{x\}$ is open and thus the collection $\{\{x\}:x\in E\}$ is an open cover of $E$, which admits a finite subcover. Hence, we have:

$E\subset \{x_1,x_2,...x_n\}$,

contradicting that $E$ is infinite. $◻$.

1 Lemma If $K$ is compact, then every ultrafilter on $K$ converges. (the Bourbaki compact property)
Proof: Let $ℱ$ be an ultrafilter on $K$. Let $E=\{\overline{F}:F\in ℱ$. Then by Lemma 1.something $E$ has the finite intersection property. Since $K$ is compact, it now follows that $\cap E$ contains a point $x$. Then one can show that $ℱ$ converges $x$. $◻$.

Separation axioms

We say a topological space is Hausdorff if two distinct points are covered by disjoint open sets. This definition can be strengthened.

1 Theorem A topological space $X$ is Hausdorff if and only if every pair of disjoint compact subsets of $X$ can be covered by two disjoint open sets.
Proof: Let $K_1,K_2\subset X$ be compact and disjoint, and $x\in K_1$ be fixed. For each $y\in K_2$ we can find disjoint open sets $A(y)$ and $B(y)$ such that $x\in A(y)$ and $y\in B(y)$. Since $K_2$ is compact, there is a finite sequence $y_1,y_2,...,y_n\in K_2$ such that:

$K_2\subset B(y_1)\cup B(y_2)...B(y_n)$.

Let $A(x)=A(y_1)\cap A(y_2)\cap ...A(y_n)$, and $B=B(y_1)\cup B(y_2)...B(y_n)$. Since $A(x)$ is disjoint from any of $B(y_1)...B(y_n)$, $A(x)$ and $B$ are disjoint. $A(x)$ is open since it is a finite intersection of open sets. Since $K_1$ is compact, there is a finite sequence $x_1,...,x_n\in K_1$ such that:

$K_1\subset A(x_1)\cup A(x_2)...A(x_n)\subset B^c\subset {K_2}^c$. $◻$

1 Corollary Every compact subset of a Hausdorff space is closed.

1 Lemma If $T$ is Hausdorff, then a filter $ℱ$ converges to at most one point.
Proof: Suppose $ℱ$ converges to two distinct points $x$ and $y$. By the separation axioms, we can find disjoint subjects $A$ and $B$ such that $x\in A$ and $y\in B$. Since convergence, $\{A,B\}\subset ℱ$. But this then implies that $A\cap B=0\in ℱ$, a contradiction. $◻$

1 Lemma Let $f$ be continuous on a set $E$. A filter $ℱ$ on $E$ converges to $x$ if and only if $f(ℱ)$ converges to $f(x)$.
Proof: Suppose $ℱ\to x$. This means that we have: $x\in A\Rightarrow E\in ℱ$. Then since the continuity means

1 Theorem Let $X$ be a topological space. The following are equivalent:

• (i) For each $x\in X$, if $y\ne x$, then there is an open set containing $y$ but not $x$.
• (ii) Every finite subset of $X$ is closed.
  

Proof: For each $x\in X$ the compliment of the singleton $\{x\}$ is the union of open sets that does not contain $x$. Thus, (i) $\Rightarrow$ (ii). For each $x\in X$, if $y\ne x$ and (ii) is true, then the compliment of $\{x\}$ is an open open set that satisfies the condition in (i). $◻$

A graph of a function $f$ is the set consisting of ordered pairs $(x,f(x))$ for all $x\in$ the domain of $f$. In the set-theoretic view, of course this set is $f$. But since we usually do not see a function as a set, the notion is often handy to use.

1 Lemma Let $f:X\to Y$ be continuous. If $Y$ satisfies the Hausdorff separation axiom, then the graph of $f$ is closed. Conversely, if the graph of $f$ is closed and $f$ is injective, then $X$ is Hausdorff.
Proof: Let $E$ be the complement of the graph of $f$. If $(x,y)\in E$, then since $f(x)\ne y$ the separation axiom says that there are $A$ and $B$, open, disjoint and such that $f(x)\in A$ and $y\in B$. It follows: $(x,y)\in f^{-1}(A)\times B\subset E$ since there is no point $z$ such that $z\in f^{-1}(A)$ and $f(z)\in B$ and the continuity says that $f^{-1}(A)$ is open. Conversely, let $(x,y)\in X$ with $x\ne y$ be given. Since $f$ is one-to-one by hypothesis $f(x)\ne f(y)$ and so $(x,f(y))\in E$, which is still the complement of the graph of $f$. If ${\pi }_1$ and ${\pi }_2$ are canonical projections, it then follows: $x\in {{\pi }_1}^{-1}(E)$, which is open by the continuity of ${\pi }_1$ and $y\in f^{-1}({{\pi }_2}^{-1}(E))$, which is again open by the continuity of ${\pi }_2$ and $f$. The two neighborhoods are disjoint by definition. $◻$

For example, the lemma shows that a topological space is Hausdorff if and only if the identity map on the space has the closed graph.

1 Lemma Let $ℱ$ be a family of functions from $X$ to a Hausdorff space $Y$. Let $\mathrm{\Gamma }$ be the union of $f^{-1}(G)$ taken all over open sets $G$ of $Y$ and $f\in ℱ$. Then $\mathrm{\Gamma }$ satisfies the Hausdorff separation axiom if and only if for each $x,y\in X$ with $x\ne y$, there is some $f\in ℱ$ such that $f(x)\ne f(y)$. Proof: First suppose the separation axiom. Then we can find two disjoint sets $A,B\in \mathrm{\Gamma }$ such that $x\in A$ and $y\in B$. Then since by definition, there is some function $f$ and sets $G_1$ and $G_2$ such that $A=f^{-1}(G_1)$ and $B=f^{-1}(G_2)$. Thus, $f(x)\ne f(y)$. Conversely, suppose the family $ℱ$ separates points in $X$. Then by the separation axiom there are disjoint open open sets $A$ and $B$ such that $x\in A$ and $y\in B$. Then by the definition of $\mathrm{\Gamma }$ $f^{-1}(A)$ and $f^{-1}(B)$ are disjoint and both open. $◻$

1 Lemma A topological space $T$ is Hausdorff if and only if for every point $p\in T$ the set $\{p\}$ is the intersection of all of its closed neighborhoods.

Zorn's Lemma

If $P$ is a collection of sets, then we say an $a$ is maximal in $P$ if $a\subset b\in P$ implies $a=b$.

1 Theorem The following are equivalent.

• Axiom of Choice.
• Zorn's Lemma: If $P$ is a nonempty collection of sets, and if every $Q\subset P$ that is totally ordered by $\subset$ contains an element $a$ such that $b\subset a$ for all $b\in Q$, then $P$ has a maximal element.

Proof: To show the converse, let $S$ be a collection of nonempty sets and $P$ be the family of all functions $f$ such that $f$ is a choice function on some $Z\subset S$. Now, Zorn's Lemma applied to $P$ gives a choice function on $S$.

Notes

Suppose a sequence of sets $E_j$. The supremum, or soup for short, of $E_j$, denoted by $\bigcup _{j=1}^{\mathrm{\infty }}$, is the smallest among sets containing all $E_j$. Similarly, the infinitum, or infu for short, of $E_j$, denoted by $\bigcap _{j=1}^{\mathrm{\infty }}$, is the largest among sets that are contained in every $E_j$. Both the supremum and the infinitum are existent and unique; since the axiomatic set theory tells us that the countable union or intersection of sets is again a set.

Now, we have the sequence of sups $\bigcup _{j=1}^{\mathrm{\infty }},\bigcup _{j=2}^{\mathrm{\infty }},\bigcup _{j=3}^{\mathrm{\infty }},...$, which is decreasing, and, similarly, the sequence of infus, $\bigcap _{j=1}^{\mathrm{\infty }},\bigcap _{j=2}^{\mathrm{\infty }},\bigcap _{j=3}^{\mathrm{\infty }},...$, which is increasing. We then define

$\underset{j\to \mathrm{\infty }}{\mathrm{lim\; sup}}{E}_j=\bigcap _{j=1}^{\mathrm{\infty }}\bigcup _{k=j}^{\mathrm{\infty }}{E}_k$

$\underset{j\to \mathrm{\infty }}{\mathrm{lim\; inf}}{E}_j=\bigcup _{j=1}^{\mathrm{\infty }}\bigcap _{k=j}^{\mathrm{\infty }}{E}_k$

and call them the limit superior and limit inferior, or colloquially lim-soup and lim-infu, of the sequence $\{E_j\}$, respectively.

When $\mathrm{lim\; sup}$ and $\mathrm{lim\; inf}$ coincide in the same set, we call it the the limit of a sequence. Obviously, the limit is always unique by the definition.

We say a sequence converges when one can always find its "tail" (i.e., the subsequence obtained after only finite number of terms are removed) in every open set containing the limit, which recall in the topological setting is a set.

1.7 Theorem There exists an exhaustion by compact sets in an open subset $\mathrm{\Omega }$ of ${ℝ}^n$; that is, a sequence of compact subsets $K_j$ of $\mathrm{\Omega }$ such that: for every $j$

• (a) $K_j\subset \text{int}{K}_{j+1}$.
• (b) If $K_j\subset \mathrm{\Omega }$ is compact, then $K\subset K_j$ for some $K_j$.
• (c) $K_j={\hat{K}}_j$.

Proof: Let {$G_{\alpha }$} be a countable basis of $\mathrm{\Omega }$. Let $K_1$ be some closed ball in $\mathrm{\Omega }$. It satisfies (c) since it is geometrically convex which implies holomorphical convexity. Suppose we have made $K_1,K_2,...K_j$ that satisfy (a) - (c). Since {$G_{\alpha }$} is a basis, we find the union $G_j$ of some open sets $G_{\alpha }$ such that $K_j\subset G_j$. Since $G_j$ has a compact closure, let $K_{j+1}=\widehat{{\overline{G}}_j}$. Then $K_{j+1}$ satisfies (a) - (c). The theorem follows after invoking inductive proof.

Axiom (Fundamental axiom of analysis) Every increasing sequence which is bounded above has a limit. (Körner )

I quote this: Everything which depends on the fundamental axiom is analysis, everything else is mere algebra (Körner)

1 Theorem For each pair $x,y\in E$, there exists an open set containing $x$ but not $y$. Then every finite subset of $E$ is closed.
Proof: Since the finite union of closed sets is closed, it suffices to show that the singleton $\{x\}$ is closed for every $x\in E$. Let $x\in E$. By hypothesis, for each $y\in E\ne x$, we can find an open set $G(y)\subset E\mathrm{\setminus }\{x\}$ containing $y$. Since we have:

$E\mathrm{\setminus }\{x\}\subset \bigcup _{y\ne x}G(y)\subset E\mathrm{\setminus }\{x\}$,

the compliment of $\{x\}$ in $E$ is open. $◻$

Historically, topologists are interested in finding the best topology for a set. The search seemed to fail to reach anywhere, however. The nicest and most familiar way to induce a topology is defining a distance function, after which a metric space results in. Since defining topology in different ways may end up with the same topology, much attention had long been paid to answer if a given topological space is metrizable; i.e., there is a way to define a distance function to get the same topology. This is called metrizable problem.

A boolean algebra is a triple $(B,+,-)$ consisting of a nonempty set B and operators $+,-$, satisfying the following axioms:

• (i) $a+b=b+a$
• (ii) $a+(b+c)=(a+b)+c$
• (iii) $-(-(a+b)+-(a+-b))=a$ (Robbins Equation)

1 Theorem (McCune's computer theorem) A boolean algebra B has the following properties:

• (1) $a+b=b+a$ and $a\cdot b=b\cdot a$ (commutative laws)
• (2) $a+(b+c)=(a+b)+c$ and $a\cdot (b\cdot c)=(a\cdot b)\cdot c$ (associative laws)
• (3) $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ and <$a+(b\cdot c)=(a+b)\cdot (a+c)$ (distributive laws)
• (4) $a\cdot (a+b)=a=a+(a\cdot b)$ (absorption law)
• (5) $-(a+b)=a\cdot b$ and $-(a\cdot b)=a+b$ (De Morgan's Laws)
• (6) There exists 0 and 1 in B such that $0\cdot a=0,1+a=1and0+a=a=1\cdot a$ (existence of identities)

Proof: For the proof which uses the computer see [3].

Two important examples of boolean algebras are power sets and a set of sentences. Indeed, let $B$ be the power set of a set E with $+$ being $\cup$, $\cdot$ being $\cap$ and $-$ being the complement with respect to $E$. Then that (i) and (ii) hold is trivial and (iii) can be shown from the definition of sets. It then follows in this $B$ $0$ is the empty set and $1$ is $E$.

References follow:

• Steen, Lynn Arthur and Seebach, J. Arthur, Jr. Counterexamples in Topology. New York, NY: Springer-Verlag, 1978. Second Edition.
• Murdeshwar, Mangesh G. General Topology. Wiley Eastern Limited 1983.