Fundamentals of Set-Theoretic Topology

Why would one want to generalize notions such as convergence and continuity to a setting even more abstract than metric spaces?

Topology deals with the relative position of objects to each other and their features. It is not about their concrete length, volume, and so on. Hence, topological features will not change if continuous transformations are applied to these objects. That is, topological features are preserved under stretching, squeezing, bending, and so on but they are not preserved under non-continuous transformations such as tearing apart, cutting and so on. That is, objects such as a circle, a rectangle and a triangle are from a topological point of view homeomorphic even though the shapes are geometrically rather different.

What features are therefore of interest such that it is worth studying topology?

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Assume that A is a closed curve, i.e. a circle, a rectangle, a triangle or something like we can see in the graph above. As long as we transform a shape A in a continuous fashion, the relative positions of the points x\in A and x\in A^C will be similar: for instance, points that have been inside of A will still be inside after the continuous transformation. Points that have been on the boundary will still be on the boundary.

Hence, the generalization of continuity and the concept of convergence are the two most characterizing features in topological spaces.

However, convergence and continuity in a metric space (X, d) were based on a notion of a distance function d(x,y) for points x,y \in X. Set-theoretic topology generalizes the features of topological metric space and ought to be based on an axiomatized notion of “closeness”.

This post is based on the literature [1] to [4].

Let (X, \mathcal{T}) — short X — be a topological space in this post.

Topology on a Set

The term “topology on a set” is based on an axiomatic description of so-called “open sets” with respect to some set-theoretic operators. It will turn out, that a topology is a set that has just enough structure to meaningful speak of convergence and continuous functions on it.

Definition 1.1 (Topological Space)
A topological space is a pair (X, \mathcal{T}), where X is a set and \mathcal{T} is a family of subsets that satisfies

(i) \emptyset, X \in \mathcal{T};
(ii) \bigcup_{i\in I}{O_i}\in \mathcal{T} if O_i \in \mathcal{T} for an arbitrary index set i\in I;
(iii) \bigcap_{i\in I}{O_i}\in \mathcal{T} if O_i \in \mathcal{T} for a finite index set I.


The following video provides a rather unorthodox way of thinking about a topology. However, it might help to get a heuristic for topological spaces. It also mentions the connection between metrics and topologies.

Topology vs. “a” Topology by PBS Infinite Series

Example 1.1 (Topologies)

(a) Let X be a set and \mathcal{T}_0=\{\emptyset, X\} then \mathcal{T}_0 is the so-called trivial, chaotic or indiscrete topology.

(b) The power set 2^X of a set X is the so-called discrete topology. In this topology every subset is open.

(c) There are four topologies on the set X:=\{0,1\}, i.e. \mathcal{T}_0:=\{\emptyset, X\}, \mathcal{T}_1:=\{\emptyset, X, \{0\}\}, \mathcal{T}_2:=\{\emptyset, X, \{1\}\}, \mathcal{T}_3:=\{\emptyset, X, \{0\}, \{1\} \} = 2^X.

(d) Let (X, \mathcal{T}) be a topological space, and let Y\subseteq X. The relative topology on Y (or the topology inherited from X) is the collection

    \begin{align*} \mathcal{T}|_{Y}:= \{ Y \cap U: \ U\in \mathcal{T} \} \end{align*}

of subsets of Y. It is clearly a topology on Y. The space (Y,\mathcal{T}|_{Y}) is then called a subspace of X.

(e) Let X be a non-empty infinite set and \mathcal{T}:=\{A \subseteq X | A=\emptyset \text{ or } X \setminus A \text{ finite} \ \} be the family of open sets. Then (X,  \mathcal{T}) is the so-called finite complement topology. First note, that an element A\in \mathcal{T}, A\neq \emptyset is of infinite cardinality. Think about what happens if we remove finitely many elements from an infinite set.

Apparently, X\in \mathcal{T} since X \setminus X = \emptyset can be considered as finite. \emptyset \in \mathcal{T} due to the definition of the set \mathcal{T} even though X \setminus \emptyset = X is not finite.

The union of arbitrary open as well as the intersection of a finitely many open sets are open again. The corresponding proof employs De Morgan’s Laws.


According to Proposition 1.1 – Fundamentals of Topology & Metric Spaces, the set \mathcal{O} of all open sets in a metric space (X, d) complies with the definition of a topology.

Definition 1.2 (Induced & Equivalent Topology)
A topology (X,\mathcal{T}_d) induced by a metric space (X, d) is defined as the set \mathcal{O} of all open sets in X.

Two metrics d_1 and d_2 on the same basic set X are called topologically equivalent if \mathcal{T}_{d_1}= \mathcal{T}_{d_2}.


Let us consider illustrative examples.

Example 1.2 (Metrizable and Equivalent Topologies)

(a) If (X, d) is a metric space and \mathcal{T}_d:= \mathcal{O} is the set of all open sets, then \mathcal{T}_d is a topology. The topology \mathcal{T}_d does not depend on the particular metric d since the proof of the referred Proposition 1.1 can also be done using the term neighborhood instead of the distance function. Hence, any metric on X equivalent to d yields the same topology. Topological spaces of this kind are called metrizable.

(b) Let us consider X\subseteq \mathbb{R}^n equipped with the natural topology by taking the topology \mathcal{T}_{d_E} induced by the Euclidean metric

    \begin{align*} d_E(x,y) = \sqrt{\sum_{i=1}^{n}{(x_i - y_i)^2} } \end{align*}

with x=(x_1, \ldots, x_n) and y=(y_1, \ldots, y_n). Instead of using the Euclidean metric, we could also employ the following distance functions:

    \begin{align*} d_M(x,y) &= \max\{|x_i - y_i| \ : i=1, \ldots, n \}, \\ d_S(x,y) &= \sum_{i=1}^{n}{(|x_i - y_i|}. \end{align*}

All three induced topologies would be equivalent, i.e. \mathcal{T}_E =  \mathcal{T}_M = \mathcal{T}_S since

    \begin{align*} \frac{1}{n} d_E(x,y) \leq  d_M(x,y) \leq d_E(x,y) \leq d_S(x,y) \end{align*}

for all x,y\in X. The corresponding unit open balls centered at (0,0)\in \mathbb{R}^2 are illustrated as follows.

An open ball d(x,0)< 1 with x\in \mathbb{R}^2 as shown in Fig. 1 is actually a set of points where each point has a distance of max. 1 to the origin (0,0). This, however, is nothing but the corresponding norm ||x||. For instance, the point x=(0.5,0.6) is not an element of the unit ball induced by d_S since 0.5+0.6>1. However, the same point x is element of the unit balls induced by d_E and d_M.

From a topological point of view the shapes in Fig. 1 are all equivalent.


Definition 1.3 (Closed Sets)
A set A is called closed in X if A^C = X\setminus A is open in X.


Example 1.3 (Closed Sets)
(a) Let (X, d) b a metric space. Then A is closed in (X, \mathcal{T}_d) if and only if A is closed in (X, d).

(b) The sets \{X, \emptyset\} are not only open but also closed for any topological space (X,\mathcal{T}) since \emptyset = X \setminus X and X = X \setminus \emptyset. The topological space is indiscrete if and only if these two sets are the only closed sets in (X,\mathcal{T}).

(c) The topology (X, \mathcal{T}) is discrete if and only if every subset A \subseteq X is closed. This can be seen by A=X\setminus (X\setminus A) = (A^C)^C.

(d) The subset [a,b] of \mathbb{R} is closed because its complement \mathbb{R} \setminus [a, b] = (-\infty, a) \cup (b, \infty) is open. Similarly, [a, +\infty) is closed, because its complement (-\infty, a) is open. The subsets [a, b) of \mathbb{R} is neither open nor closed.

(e) In the finite complement topology on an infinite set X, the closed sets consists of X itself and all finite subsets of X.


Proposition 1.1: (Characterization of Closed Sets)
(1) The set \mathcal{S} of all closed sets of (X, \mathcal{T}) complies with the following conditions:
(i) \emptyset \in \mathcal{S} and X\in \mathcal{S}.
(ii) A, B \in \mathcal{S} implies A \cup B\in \mathcal{S}.
(iii) \mathcal{A} \subseteq \mathcal{S} implies \bigcap_{A\in \mathcal{A}}{A}\in \mathcal{S}.

(2) Let \mathcal{A} be a family of sets that complies with (i), (ii) and (iii) then there exists a topology (X, \mathcal{T}), such that \mathcal{S} is the set of all closed sets in (X, \mathcal{T}).

Proof. (1) This follows directly from the definitions and applying the rules X \setminus X=\emptyset, X\setminus \emptyset=X, X \setminus (A \cup B) = (X\setminus A) \cap (X \setminus B) as well as (X \setminus \bigcap_{A\in \mathcal{A}}{A} = \bigcup_{A\in \mathcal{A}}{X\setminus A}.

(2) The family of closed sets \mathcal{S} fully determines the topology \mathcal{T} on the same basic set X since \mathcal{S}=\{ A | \ (X\setminus A)\in \mathcal{T} \}. Its existence follows from the fact that \mathcal{S} actually is a topology but this is clear given (1).


The family of closed sets of a topology could also be used to define a topological space, i.e. the set of all closed sets contains exactly the same information as the set of all open sets that actually define the topology.

Definition 1.4 (Interior & Closure)
Let (X, \mathcal{T}) be a topology. The interior of A\subset X is defined as the union of all open sets contained in A, i.e.

    \begin{align*} \text{Int} A = \bigcup_{T \subseteq A}{T}, \quad T\in \mathcal{T}. \end{align*}

The closure of A is defined as the intersection of all closed sets containing A, i.e.

    \begin{align*} \text{Cl} A = \bigcap_{A \subseteq S}{S}, \quad S\in \mathcal{S}. \end{align*}

A neighborhood A of a point x_0 in a topological space is any set containing x_0 as well as an open set B\in \mathcal{T} of x_0, i.e. \{x_0\} \subseteq B \subseteq A.


A set A is open in a Euclidean metric space if and only if for every x_0\in A an open ball B(x_0, r_0) exists such that B(x_0, r_0) \subseteq A. Let us therefore consider the situation in the real plane \mathbb{R}^2 employing the topology induced by the Euclidean metric. The open balls B(x_1, r_1) and B(x_2, r_2) are both contained in A and thus a part of the interior IntA. Thereby, B(x_i, r_i) can be considered as neighborhoods of x_i with i=1,2. A is a neighborhood of x_1, x_2.

The open ball B(x_0, r_0) is not fully contained in A, which is why A is not a neighborhood of x_0 and also not an element of the interior of A.

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The dual situation of the closure ClA can also be seen in the figure above by considering A^C. An important type of points in such situations is the following.

Definition 1.2 (Boundary Point)
Let (X, \mathcal{T}), A\subseteq X and x\in X. Then, x is called boundary point for A if

(1)   \begin{align*} B(x, \epsilon) &\cap A \neq \emptyset \quad \text{ and } \\ B(x, \epsilon) &\cap A^C \neq \emptyset \\ \end{align*}

for all \epsilon>0. The set of all boundary points is denoted by \partial A.


Boundary points are just points on the boundary between the set A and the surrounding basis set X (i.e. A^C) of the metric space (X, d).

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\partial A can be contained in A or in A^C. If all boundary points \partial A are outside of A, i.e. if A \cap \partial A = \emptyset, then it is an open set.

Considering Example 1.2 (a), we could ask whether every topological space is metrizable?

The answer is no, and the root-cause is that topological spaces have different types of separation properties.

Separation Properties

A metric enables us to separate points in a metric space since any two distinct points have a strictly positive distance. In general topological spaces, separating points from each other is more subtle.

Hausdorff Space

Hausdorff spaces and the Hausdorff condition are named after Felix Hausdorff, one of the founders of topology. Let us first check out the formal definition.

Definition 2.1: (Hausdorff Space, T_2 Spaces)
A topological space X is a Hausdorff or T_2-space if, for any pair of distinct points x, y\in X, x\neq y there are disjoint open sets U, V\in \mathcal{T} with x\in U, y\in V and U\cap V = \emptyset.


Every Euclidean space is Hausdorff since we can use the Euclidean metric to separate two distinct points. The following video outlines the Hausdorff condition and it provides a simple example of a Hausdorff space.

Hausdorff Condition incl. an Example and Counterexample by DanielChanMaths

Example 2.1 (Metric Space is Hausdorff)
Let (X, d) be a metric space, and let x,y\in X be such that x\neq y. It follows that \epsilon:=0.5 \cdot d(x,y)>0. Let U:=B(x, \epsilon) and V:=B(y, \epsilon). Then, X is Hausdorff since U\cap V=\emptyset.


In a Hausdorff space, distinct points can be separated by open sets.

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The situation in X:=\mathbb{R}^2 along with the topology implied by the Euclidean metric is illustrated in the graph above. For two distinct points x_1, x_2\in X, we take half (or less) the distance to define \epsilon to come up with two distinct open balls, that can also be seen as disjoint neighborhoods.

Proposition 2.1: (Subset of T_2-spaces)
Let (X, \mathcal{T}) be a topological Hausdorff space. Then, each subset A\subseteq X of a Hausdorff space is Hausdorff.

Proof. Let x\neq y be in X. The space being Hausdorff, let U_x and V_y be the two open separating sets as required in Definition 1.1. Then U:=U_x \cap A as well as V:=V_y \cap A are open since the difference of two open sets is open. In addition, x\in U and y\in V.


T_0– and T_1– Spaces

The following separation properties are weaker than the Hausdorff (T_2-) condition. This is also indicated by the index of the corresponding names of the separation axioms (from T_0 to T_2).

Definition 2.2. (T_0 Space)
A topological space (X, \mathcal{T}) is called a T_0– or Kolmogorov space if, for any x,y\in X with x\neq y, there is an open set U\subseteq X with x\in U and y\notin U or y\in U and x\notin U.


The most striking difference between a Hausdorff / T_2– and T_0-space is that only one open set U, that contains only one of two distinct points, is required to fulfill the definition of a T_0-space. Apparently, every T_2-space is also a T_0-space.

Example 2.2:
(a) Let X=\{x,y\} be any set with at least two elements equipped with the so-called chaotic topology \mathcal{T}:=\{\emptyset, X\}. Then, there is no open U\in \mathcal{T} that separates the two distinct elements. Hence, this topology is not a T_0-space and definitely also not a Hausdorff space. Hence, it is also not metrizable.

(b) Let X=\{x,y\} be any set with the discrete topology \mathcal{T}:=2^X. Then, \{x\} \in \mathcal{T} separates the two elements x and y.


In a T_1-space two open sets are required to separate two distinct points, however, the two sets don’t need to be disjoint.

Definition 2.3. (T_1 Space)
A topological space (X, \mathcal{T}) is called a T_1-space if, for any x,y\in X with x\neq y, there are open sets U, V\subseteq X with x\in U and y\notin U and y\in U and x\notin U.


The main difference between a T_2– and a T_1-space is that the two required open sets do not have to be disjoint. However, one open set only contains one of the two distinct points.

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Hence, every T_1-space is also a T_0-space. Just take one of the two open sets of the T_1-space and it fulfills all requirements of a T_0-space.

Proposition 2.2: (Characterization of T_1-spaces)
(a) Let (X, \mathcal{T}) be a topological space. Then, X is a T_1-space if and only if \{x\} is a closed set for each x\in X;
(b) Each Hausdorff space is a T_1-space.

Proof. (a) Suppose X is a T_1-space, and let x\in X. For any y\in X with y\neq x, there is an open subset U_y of X with y\in U_y, but x\notin U_y. It follows that X\setminus \{x\}=\bigcup{ \{U_y: y\in X, x\neq y\} }.

Conversely, suppose that all singleton subsets of X are closed, and let x,y\in X be such that x\neq y. Then, V:= X\setminus \{x\} and U:= X\setminus \{y\} fulfill the requirements of a T_1-space.

(b) Let x be a given point. By assumption, each y\neq x belongs to an open set U_y such that x\notin G_y. Consequently, X \setminus \{x\} = \bigcup_{y\neq x}{U_y}. Thus, X\setminus \{x\} is open, and \{x\} is closed.


Convergent Sequences

One of the key features of topological spaces is the generalization of the convergence concept.

A sequence in a (metric) space X is a function x_n:\mathbb{N} \rightarrow X that we also denote by (x_n), in particular, if we want to refer to the elements of the sequence. Given a sequence x_n in a metric space, a sub-sequence is the restriction of x_n to an infinite subset S \subseteq \mathbb{N}. If we exhibit S as n_1<n_2<\ldots <n_k<\ldots, then we write the subsequence (x_{n_k}).

We say that a sequence (x_n) converges to x\in X if given \epsilon>0, there exist N\in \mathbb{N} such that for all n\in \mathbb{N}, we have x_n\in B(x, \epsilon).

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In other words, for all n\geq N, we have x_n\in B(x, \epsilon) as illustrated in Figure above. The finitely many elements x_1, \ldots, x_{n-1} of the sequence (x_n) are, however, not contained in B(x, \epsilon). We take this property to define what a convergent sequence in a topological space is.

Definition 3.1. (Convergent Sequence)
Let (X, \mathcal{T}_d) be a topological space. A sequence (x_n) converges to x\in X if \mathcal{T}_d, the set \{k \ | \ x_k\notin U\} is finite for any open set U\ni x.

The point x\in X is then called the limit of the sequence (x_n) and we denote it by x = \lim_{n\rightarrow \infty}{x_n} or by x_n \rightarrow x.


Note that the set of \{k \ | \ x_k\in U\} = \{k \ | \ x_k\notin U\}^C is infinite for all open sets U\ni x while (at the same time) the set \{k \ | \ x_k\notin U\} is finite. Both sets/conditions matter in this situation as we will see further below!

Lemma 3.1. (Limit of a sequence is unique)
The limit x of a convergent sequence (x_n)\rightarrow x in a Hausdorff space (X, d) is unique.

Proof. Assume that this is not the case and x_n \rightarrow x as well as x_n \rightarrow y with x\neq y holds true. A metric space is Hausdorff, that is, we find two disjoint open balls B(x, r) and B(y, r). Given that x and y are the limit points almost all elements must lie in the disjoint balls, which contradicts the initial assumption of x\neq y.


Lemma 1.1 is false in arbitrary topological spaces.

Every x\in X of a topological space (X, \mathcal{T}) is the limit of a certain sequence (x_n). Apparently, we could simply use the constant sequence x_1=x, x_2=x, \ldots or we could define x_k:=x for all k\geq N, N\in \mathbb{N}. This fact should be also considered in the following examples.

Example 3.1:
(a) Let (X, \mathcal{T}) be the discrete topology with \mathcal{T}=2^X. Further, let x_n \rightarrow x. Recall that in this topology every set is open by definition. Hence, also \{x\} is an open set that must be contained in all other (open) supersets. Hence, the set \{k \ | \ x_k\neq x\} has to be finite and \{k \ | \ x_k= x\} has to be infinite.

(b) Let (X, \mathcal{T}) be the indiscrete topology with \mathcal{T}=\{\emptyset, X\}. Further, let x_n \rightarrow x. Since U=X is the only set that contains x\in X, the set \{k \ | \ x_k \notin U\} has to be finite for every sequence (x_n). Hence, every sequence converges to every point in x\in X.


Closely related to converging sequences and their limits are accumulation points.

Definition 3.2. (Accumulation Point)
An element x of a sequence (x_n) is called accumulation point (sometimes also cluster or limit point) if \{k \ | \ x_k\in U\} is infinite for every open set U of x.


The subtle but important difference between an accumulation point and a limit is that the complement set of \{k \ | \ x_k\in U\} can also be infinite. Let us consider a simple example.

Example 3.2:
Let us consider the sequence (x_n):= \{(-1)^n \ | \ n\in \mathbb{N}\} in the topology induced by the Euclidean space (X=\mathbb{R}, d_E) on the real line. There are two accumulation points \pm 1 but no limit of the sequence. Note that the sequence is alternating between +1 and -1, such that \{k \ | \ (-1)^{2k} = 1\} and \{k \ | \ (-1)^{2k} = -1\} are both infinite but disjoint to each other. In addition, the set \{k \ | \ x_k\in U\} for an open set U of x=\pm 1 are both infinite.


A sub-sequence (x_{n_k}) of a convergent sequence (x_n) converges to the same limit x. This is evident since if the condition of a convergent sequence is fulfilled for all elements x_n, n\in \mathbb{N} of the series (x_n) . Hence, the condition is also fulfilled for a subset (x_{n_k}) \subseteq (x_n) that represents the sub-sequence.

Due to the fact that the finiteness of \{k \ | \ x_k\notin U\} implies the infiniteness of \{k \ | \ x_k\in U\} in \mathbb{N} every limit is an accumulation point. The converse is not true as we can see in Example 3.2.

Theorem 3.1 (Convergence in Topological Spaces)
Let (X, \mathcal{T}) be a topological space.
(i) Every limit of a convergent sequence is also the limit of any sub-sequence.
(ii) Every accumulation point of any sub-sequence (x_{n_k}) is also an accumulation point of (x_n).

Proof. (i) If x_n \rightarrow x then \{k | x_k\notin U\} is finite for all open U of x. In particular, this holds true for any sub-sequence (x_{n_k})\subseteq (x_n) and thus x_{n_k} \rightarrow x.

(ii) Let p\in X be an accumulation point of the sub-sequence, i.e. the set \{k \ | \ x_{k}\in U\} is infinite for every open set U of x. Since the sub-sequence is only a subset of the element of the sequence, the assertion follows directly.


Completely Regular Spaces

Our next separation axiom has a somewhat different flavor; it is not defined in terms of topology, but via continuous functions.

Definition 4.1: (Completely Regular Space)
Let (X, \mathcal{T}) be a T_1-space. Then X is called completely regular if and only if, for every point x\in X and every closed set F\subseteq X, x\notin F, there is a function f\in C(X, \mathbb{R}) with f(X) \subseteq [0,1], f(x)=1, and f|_{F}=0.


A completely regular space X posses a family of corresponding functions f:X\rightarrow [0,1] with x\mapsto f(x)=1 and f(F)=0 for all closed F\subseteq X, x\notin F. Let us consider an example.

Example 4.1: (Completly Regular Metric Space)
Let (X, d) be a metric space, let x_0\in X, and let F\subseteq X be closed such that x_0\notin F. To avoid triviality, suppose F\neq \emptyset. Then, define

    \begin{align*} g: X \rightarrow \mathbb{R}, \quad x \mapsto d(x, F) \end{align*}


Definition 4.2: (Locally Compact)
A topological space X is locally compact if every point x\in X has a neighborhood U whose closure K:=\overline{U} is compact.


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Example 4.2: (Locally Compact \mathbb{R}^n)
The topology (X=\mathbb{R}^n, \mathcal{T}=\mathcal{O}) where \mathcal{O} are the open sets of the Euclidean metric is locally compact.

According to the Heine-Borel Theorem, C\subseteq X is closed and bounded if and only if C is compact. Every point x\in \mathbb{R}^n posses infinitely many open sets B(x, \epsilon), \epsilon>0 and corresponding closed balls \overline{B(x, \epsilon)}. The closed balls are bounded by definition (i.e. \epsilon\in \mathbb{R}) and they are also closed since its complement \mathbb{R}^n \setminus \overline{B(x, \epsilon)} is open. Refer to Example 1.2 as well as Example 1.3 (d) in this post for further details. Thus, the set \overline{B(x, \epsilon)} is compact and \mathbb{R}^n locally compact.


The compactness (i.e. boundedness and closedness) is needed in combination with continuous functions as we will see further below.

A subset A\subseteq X of a locally compact space is bounded if there exists a compact set C such that A \subseteq C.

For any topological space X, we denote by \mathcal{F} the class of all real-valued continuous functions f such that 0\leq f(x)\leq 1 for all x\in X.

Theorem 4.1:
If C is a compact set and U and V are open sets such that C \subseteq U\cup V, then there exist compact sets D and E such that D\subseteq U, E\subseteq V, and C=D\sup E.

Proof. Since C\setminus U and C\setminus V are disjoint compact sets, there exist two disjoint open sets U^* and V^*, such that C \setminus U \subset U^* and C \setminus V \subset V^*. We write D=C \setminus U^* and E=C \setminus V^*. It is easy to verify that D \subset U, E \subset V, and that D and E are compact. Since U^*\cap V^* = \emptyset, we have D \cup E = (C\setminus U^*) \cup (C\setminus V^*) = C \setminus (U^*\cap V^*)=C.


Theorem 1.1 tells us that if we have a two-element open cover of a compact set C then we can find a two-element closed cover of C.

Another important theorem is the following.

Theorem 4.2:
If C is a compact set and F is a closed disjoint set, i.e. C \cap F = \emptyset. Then there exists a function f in \mathcal{F} such that f(x)=0 for x\in C and f(x)=1 for x\in F.

Proof. Since X is completely




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