Chapter 2 • Theory & Derivations

Unit 2: Topological Spaces, Bases, Subbases & Neighborhood Systems

Axiomatic foundations of general topology: definition of topological spaces, comparison of topologies (coarser and finer), classical non-metric topologies (co-finite, co-countable, discrete, indiscrete, Sierpiński), closed sets, interior, closure, boundary, Kuratowski Closure Axioms, neighborhood systems and filters, accumulation and derived points, topological bases and subbases, the Basis Criterion Theorem, and the subspace (relative) topology.

§2.1 Axiomatic Definition of Topological Spaces & Comparison of Topologies

1. The Axioms of a Topological Space

Felix Hausdorff in 1914 and Kazimierz Kuratowski in 1922 abstracted metric open sets into the general axiomatic framework of topology.

Definition 2.1 (Topological Space): Let $X$ be a non-empty set. A topology on $X$ is a collection $\mathcal{T}$ of subsets of $X$ (whose members are called open sets) satisfying:

The empty set $\emptyset$ and the whole space $X$ belong to $\mathcal{T}$:

\emptyset \in \mathcal{T} \quad \text{and} \quad X \in \mathcal{T}

The union of any arbitrary family of sets in $\mathcal{T}$ belongs to $\mathcal{T}$:

\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T} \quad \text{whenever } U_\alpha \in \mathcal{T} \text{ for all } \alpha \in I

The intersection of any finite collection of sets in $\mathcal{T}$ belongs to $\mathcal{T}$:

\bigcap_{i=1}^n U_i \in \mathcal{T} \quad \text{whenever } U_1, \dots, U_n \in \mathcal{T}

The pair $(X, \mathcal{T})$ is called a topological space.

2. Classical Non-Metric Topologies

Example 2.1 (Discrete and Indiscrete Topologies):

Discrete Topology $\mathcal{T}_{\text{disc}} = \mathcal{P}(X)$: Every subset of $X$ is open.

Indiscrete (Trivial) Topology $\mathcal{T}_{\text{ind}} = \{\emptyset, X\}$: Only $\emptyset$ and $X$ are open.

Example 2.2 (Co-finite Topology / Finite Complement Topology): Let $X$ be any set. Define:

\mathcal{T}_{\text{cof}} = \{U \subseteq X : X \setminus U \text{ is finite}\} \cup \{\emptyset\}

On an infinite set $X$, $\mathcal{T}_{\text{cof}}$ is not metrizable! Any two non-empty open sets must intersect, so disjoint open neighborhoods cannot exist for distinct points!

Example 2.3 (Co-countable Topology):

\mathcal{T}_{\text{coc}} = \{U \subseteq X : X \setminus U \text{ is countable}\} \cup \{\emptyset\}

On an uncountable set like $\mathbb{R}$, sequences converge if and only if they are eventually constant.

Example 2.4 (Sierpiński Space): On the two-point set $X = \{0, 1\}$, the topology $\mathcal{T} = \{\emptyset, \{1\}, \{0, 1\}\}$ is called the Sierpiński topology. The point $1$ is open but not closed, while $0$ is closed but not open!

3. Comparison of Topologies

Definition 2.2 (Coarser and Finer Topologies): Let $\mathcal{T}_1$ and $\mathcal{T}_2$ be two topologies on the same set $X$.

If $\mathcal{T}_1 \subseteq \mathcal{T}_2$, we say that $\mathcal{T}_1$ is coarser (or weaker, smaller) than $\mathcal{T}_2$, and that $\mathcal{T}_2$ is finer (or stronger, larger) than $\mathcal{T}_1$.

For any set $X$:

\mathcal{T}_{\text{ind}} \subseteq \mathcal{T}_{\text{cof}} \subseteq \mathcal{T}_{\text{metric}} \subseteq \mathcal{T}_{\text{disc}}

If neither $\mathcal{T}_1 \subseteq \mathcal{T}_2$ nor $\mathcal{T}_2 \subseteq \mathcal{T}_1$, the topologies are called incomparable.

§2.2 Closed Sets, Interior, Closure, Boundary & Kuratowski Axioms

1. Closed Sets, Closure and Interior

Definition 2.3 (Closed Set): A subset $F \subseteq X$ is called closed in $(X, \mathcal{T})$ if its complement $X \setminus F$ is open in $(X, \mathcal{T})$.

Theorem 2.1 (Properties of Closed Sets):

$\emptyset$ and $X$ are closed.

The intersection of an arbitrary family of closed sets is closed: $\bigcap_{\alpha \in I} F_\alpha$ is closed.

The union of a finite collection of closed sets is closed: $\bigcup_{i=1}^n F_i$ is closed.

Definition 2.4 (Interior, Closure, Boundary): Let $A \subseteq X$.

The interior of $A$, denoted $\operatorname{int}(A)$ or $A^\circ$, is the largest open set contained in $A$:

\operatorname{int}(A) = \bigcup \{U \in \mathcal{T} : U \subseteq A\}

The closure of $A$, denoted $\operatorname{cl}(A)$ or $\bar{A}$, is the smallest closed set containing $A$:

\operatorname{cl}(A) = \bigcap \{F \subseteq X : F \text{ is closed and } A \subseteq F\}

The boundary (or frontier) of $A$ is:

\partial A = \operatorname{cl}(A) \setminus \operatorname{int}(A) = \operatorname{cl}(A) \cap \operatorname{cl}(X \setminus A)

The exterior of $A$ is $\operatorname{ext}(A) = \operatorname{int}(X \setminus A) = X \setminus \operatorname{cl}(A)$.

2. Kuratowski Closure Axioms

Topological spaces can be defined entirely through the closure operator rather than open sets!

Theorem 2.2 (Kuratowski Closure Axioms, 1922): Let $X$ be a set. A closure operator is a mapping $c: \mathcal{P}(X) \to \mathcal{P}(X)$ satisfying for all $A, B \subseteq X$:

Empty Set: $c(\emptyset) = \emptyset$.

Extension: $A \subseteq c(A)$.

Additivity: $c(A \cup B) = c(A) \cup c(B)$.

Idempotence: $c(c(A)) = c(A)$.

If an operator $c$ satisfies these four axioms, there exists a unique topology $\mathcal{T}$ on $X$ such that for all $A \subseteq X$, $\operatorname{cl}(A) = c(A)$, where closed sets are precisely the fixed points: $F \text{ is closed} \iff c(F) = F$.

§2.3 Neighborhood Systems, Neighborhood Filters & Accumulation Points

1. Neighborhoods and Neighborhood Filters

Definition 2.5 (Neighborhood): Let $(X, \mathcal{T})$ be a topological space and $x \in X$. A subset $N \subseteq X$ is called a neighborhood of $x$ if there exists an open set $U \in \mathcal{T}$ such that:

x \in U \subseteq N

Note that $N$ itself does not need to be open; if $N$ is open, it is called an open neighborhood.

Theorem 2.3 (Hausdorff Neighborhood Filter Axioms): The collection $\mathcal{N}(x)$ of all neighborhoods of a point $x$ forms a filter on $X$:

Non-emptiness: $N \in \mathcal{N}(x) \implies x \in N$.

Supersets: If $N \in \mathcal{N}(x)$ and $N \subseteq M$, then $M \in \mathcal{N}(x)$.

Finite Intersections: If $N_1, N_2 \in \mathcal{N}(x)$, then $N_1 \cap N_2 \in \mathcal{N}(x)$.

Open Core: If $N \in \mathcal{N}(x)$, there exists $U \in \mathcal{N}(x)$ such that $N \in \mathcal{N}(y)$ for every $y \in U$.

2. Accumulation Points, Derived Sets & Isolated Points

Definition 2.6 (Limit / Accumulation Point): Let $A \subseteq X$. A point $x \in X$ is called an accumulation point (or limit point, cluster point) of $A$ if every neighborhood $U$ of $x$ contains at least one point of $A$ different from $x$:

(U \setminus \{x\}) \cap A \ne \emptyset, \qquad \forall U \in \mathcal{N}(x)

The set of all accumulation points of $A$ is called the derived set of $A$, denoted $A'$.

Points in $A \setminus A'$ are called isolated points of $A$.

A point $x \in X$ is an adherent point of $A$ if every neighborhood of $x$ intersects $A$: $U \cap A \ne \emptyset$.

Theorem 2.4 (Closure via Derived Set): For any subset $A \subseteq X$:

\operatorname{cl}(A) = A \cup A'

Consequently, a subset $A$ is closed if and only if it contains all of its accumulation points ($A' \subseteq A$).

§2.4 Bases and Subbases for a Topology

1. Bases for a Topology

Specifying every single open set in an infinite topology is practically impossible. Instead, topologies are defined via a smaller collection of generating sets called a basis.

Definition 2.7 (Basis for a Topology): Let $(X, \mathcal{T})$ be a topological space. A collection $\mathcal{B} \subseteq \mathcal{T}$ of open sets is called a basis for $\mathcal{T}$ if every open set $U \in \mathcal{T}$ can be expressed as a union of elements of $\mathcal{B}$:

U = \bigcup \{B \in \mathcal{B} : B \subseteq U\}

Equivalently, for every $U \in \mathcal{T}$ and every $x \in U$, there exists $B \in \mathcal{B}$ such that $x \in B \subseteq U$.

2. The Basis Criterion Theorem

When does an arbitrary family of subsets $\mathcal{B}$ form a basis for some topology on $X$?

Theorem 2.5 (Basis Criterion): A collection $\mathcal{B}$ of subsets of $X$ is a basis for a topology on $X$ if and only if:

Covering Property: $\bigcup_{B \in \mathcal{B}} B = X$ (i.e. every point $x \in X$ belongs to at least one $B \in \mathcal{B}$).

Intersection Property: For any two basis elements $B_1, B_2 \in \mathcal{B}$ and any point $x \in B_1 \cap B_2$, there exists a basis element $B_3 \in \mathcal{B}$ such that:

x \in B_3 \subseteq B_1 \cap B_2

If these two conditions hold, the topology $\mathcal{T}(\mathcal{B})$ generated by $\mathcal{B}$ consists of all arbitrary unions of sets in $\mathcal{B}$.

Proof: ($\Rightarrow$) If $\mathcal{B}$ is a basis for $\mathcal{T}$, since $X \in \mathcal{T}$, $X$ is a union of basis elements, so $\bigcup \mathcal{B} = X$. Since $B_1, B_2 \in \mathcal{T}$, their intersection $B_1 \cap B_2 \in \mathcal{T}$. Since $\mathcal{B}$ is a basis, for any $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2$.

($\Leftarrow$) Define $\mathcal{T}$ as the collection of all unions of subsets of $\mathcal{B}$.

$\emptyset$ is the empty union ($\emptyset \in \mathcal{T}$). By condition (1), $X = \bigcup_{B \in \mathcal{B}} B \in \mathcal{T}$.

Unions of sets in $\mathcal{T}$ are obviously unions of sets in $\mathcal{B}$, hence in $\mathcal{T}$.

For finite intersections: it suffices to check that if $U, V \in \mathcal{T}$, then $U \cap V \in \mathcal{T}$.

Let $x \in U \cap V$. Then $x \in B_1 \subseteq U$ and $x \in B_2 \subseteq V$ for some $B_1, B_2 \in \mathcal{B}$. By condition (2), there exists $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq B_1 \cap B_2 \subseteq U \cap V$. Then $U \cap V = \bigcup_{x \in U \cap V} B_x \in \mathcal{T}$. $\blacksquare$

3. Subbases for a Topology

Definition 2.8 (Subbasis): A collection $\mathcal{S}$ of subsets of $X$ is called a subbasis for a topology $\mathcal{T}$ if the collection of all finite intersections of elements of $\mathcal{S}$:

\mathcal{B} = \left\{ \bigcap_{i=1}^k S_i : k \in \mathbb{N}, \; S_i \in \mathcal{S} \right\}

forms a basis for $\mathcal{T}$ (with the empty intersection defined as $X$). Remark: Any arbitrary family of subsets $\mathcal{S}$ covering $X$ generates a unique topology $\mathcal{T}$ on $X$ for which $\mathcal{S}$ is a subbasis!

§2.5 Subspace Topology (Relative Topology) & Hereditary Properties

1. The Subspace Topology

Definition 2.9 (Subspace Topology): Let $(X, \mathcal{T})$ be a topological space and let $Y \subseteq X$ be an arbitrary subset. The subspace topology (or relative topology) on $Y$, denoted $\mathcal{T}_Y$, is defined by:

\mathcal{T}_Y = \{U \cap Y : U \in \mathcal{T}\}

The pair $(Y, \mathcal{T}_Y)$ is called a topological subspace of $(X, \mathcal{T})$.

2. Relative Open and Closed Sets

Proposition 2.1 (Relative Open and Closed Sets): Let $Y \subseteq X$.

A set $V \subseteq Y$ is open in $Y$ if and only if $V = U \cap Y$ for some open set $U \subseteq X$.

A set $K \subseteq Y$ is closed in $Y$ if and only if $K = F \cap Y$ for some closed set $F \subseteq X$.

If $Y$ is open in $X$, then every set open in $Y$ is also open in $X$.

If $Y$ is closed in $X$, then every set closed in $Y$ is also closed in $X$.

Theorem 2.6 (Relative Closure and Interior): For any subset $A \subseteq Y$:

The closure of $A$ in $Y$ satisfies:

\operatorname{cl}_Y(A) = \operatorname{cl}_X(A) \cap Y

The interior of $A$ in $Y$ satisfies:

\operatorname{int}_X(A) \cap Y \subseteq \operatorname{int}_Y(A)

(The inclusion can be strict: for example, the interval $[0, 1/2)$ has non-empty interior in the subspace $Y = [0, 1]$, but empty interior in $\mathbb{R}$!).

3. Hereditary Properties

Definition 2.10 (Hereditary Property): A topological property $P$ is called hereditary if whenever a space $X$ possesses property $P$, every subspace $Y \subseteq X$ also possesses property $P$. It is called weakly hereditary (or closed-hereditary) if it is inherited by all closed subspaces.

Examples of hereditary properties: Hausdorff ($T_2$), First-countability, Second-countability, Metrizability.

Examples of non-hereditary properties: Compactness (a closed interval $[0, 1]$ is compact, but its subspace $(0, 1)$ is not!).

Tiered Solved Examination Problems & Rigorous Derivations

Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.

Foundational Example 2.1: Topological Interior, Closure and Boundary in Co-finite Topology

Let $X = \mathbb{R}$ be equipped with the co-finite topology $\mathcal{T}_{\text{cof}}$, where non-empty open sets are those with finite complement. Compute the interior, closure, and boundary in $(X, \mathcal{T}_{\text{cof}})$ for each of the following subsets:

$A = \{1, 2, 3, \dots, 10\}$ (a finite set)
$B = (0, 1)$ (the open unit interval)
$C = \mathbb{Z}$ (the integers)
$D = \mathbb{Q}$ (the rational numbers)

1. Closed Sets in $(X, \mathcal{T}_{\text{cof}})$

By definition, a subset $F \subseteq \mathbb{R}$ is closed in $\mathcal{T}_{\text{cof}}$ if and only if $F = \mathbb{R}$ or $F$ is finite. Non-empty open sets $U$ are those where $\mathbb{R} \setminus U$ is finite.

2. Set $A = \{1, 2, \dots, 10\}$ (Finite Set)

Closure: Since $A$ is finite, $A$ is closed. Thus $\operatorname{cl}(A) = A = \{1, 2, \dots, 10\}$.
Interior: The only open sets in $\mathcal{T}_{\text{cof}}$ are $\emptyset$ and sets with finite complement.

Since $A$ is finite, its complement $\mathbb{R} \setminus A$ is infinite (uncountable), so $A$ cannot contain any non-empty open set! Thus $\operatorname{int}(A) = \emptyset$.

Boundary: $\partial A = \operatorname{cl}(A) \setminus \operatorname{int}(A) = A \setminus \emptyset = A = \{1, 2, \dots, 10\}$.

3. Set $B = (0, 1)$ (Infinite Set with Infinite Complement)

Closure: The only closed sets containing $B$ are closed sets of $\mathcal{T}_{\text{cof}}$.

The only closed sets are finite sets and $\mathbb{R}$ itself. Since $B$ is infinite, no finite set can contain $B$. The only closed set containing $B$ is the whole space $\mathbb{R}$! Thus $\operatorname{cl}(B) = \mathbb{R}$.

Interior: For $B$ to contain a non-empty open set $U$, $\mathbb{R} \setminus U$ must be finite.

Then $\mathbb{R} \setminus B \subseteq \mathbb{R} \setminus U$ would force $\mathbb{R} \setminus B$ to be finite. However, $\mathbb{R} \setminus (0, 1) = (-\infty, 0] \cup [1, \infty)$ is infinite! Thus $B$ contains no non-empty open set, so $\operatorname{int}(B) = \emptyset$.

Boundary: $\partial B = \operatorname{cl}(B) \setminus \operatorname{int}(B) = \mathbb{R} \setminus \emptyset = \mathbb{R}$.

4. Sets $C = \mathbb{Z}$ and $D = \mathbb{Q}$

Both $C = \mathbb{Z}$ and $D = \mathbb{Q}$ are infinite subsets with infinite complement:

$\operatorname{cl}(\mathbb{Z}) = \mathbb{R}$, $\operatorname{int}(\mathbb{Z}) = \emptyset$, $\partial \mathbb{Z} = \mathbb{R}$.
$\operatorname{cl}(\mathbb{Q}) = \mathbb{R}$, $\operatorname{int}(\mathbb{Q}) = \emptyset$, $\partial \mathbb{Q} = \mathbb{R}$.

In the co-finite topology, every infinite subset is dense ($\bar{S} = X$), and every subset with infinite complement has empty interior!

Advanced Example 2.2: The Kuratowski 14-Set Problem

In 1922, Kazimierz Kuratowski asked: given a subset $A$ of a topological space $(X, \mathcal{T})$, how many distinct sets can be formed by repeatedly applying the closure operator $c(A) = \bar{A}$ and the complement operator $k(A) = X \setminus A$?

Prove that at most $14$ distinct sets can be obtained.
Construct an explicit subset $A \subset \mathbb{R}$ under the standard Euclidean topology that achieves all $14$ distinct sets.

1. Theoretical Maximum of 14 Sets

Let $c(A) = \operatorname{cl}(A)$ and $k(A) = X \setminus A$. Note the fundamental relations:

$k(k(A)) = A$ (involution: $k^2 = \operatorname{id}$).
$c(c(A)) = c(A)$ (idempotence: $c^2 = c$).
The interior operator is $i(A) = k(c(k(A)))$. Thus $i^2 = i$.

Any sequence of operations alternates between $c$ and $k$. Consider words starting with $A$:

Words of the form $k, ck, kck, ckc, \dots$
Words of the form $c, kc, ckc, kckc, \dots$

Key Identity (Kuratowski): For any subset $A$:

$$c(k(c(k(c(k(c(A))))))) = c(k(c(A)))$$

That is:

$$c i c i(A) = c i(A)$$

Proof of Identity: Since $i(A) \subseteq A$, taking closure gives $c(i(A)) \subseteq c(A)$. Taking interior gives $i(c(i(A))) \subseteq i(c(A))$. Taking closure gives $c i c i(A) \subseteq c i(A)$. Conversely, $i(A)$ is open, so $i(A) \subseteq i(c(i(A)))$, and taking closure yields $c i(A) \subseteq c i c i(A)$. Thus:

$$c i c i(A) = c i(A)$$

Dualizing with complement gives $i c i c(A) = i c(A)$.

Because the alternating sequences of $c$ and $i$ collapse after 2 iterations: The only possible operations on $A$ are:

$A$
$c(A)$
$i c(A)$
$c i c(A)$
$i(A)$
$c i(A)$
$i c i(A)$

Applying complement $k$ to each of these $7$ sets produces at most $7$ complementary sets:

7 + 7 = 14 \text{ distinct sets!}

2. Explicit Subset of $\mathbb{R}$ Achieving 14 Sets

Consider the subset $A \subset \mathbb{R}$ defined by:

A = (0, 1) \cup (1, 2) \cup \{3\} \cup (\mathbb{Q} \cap [4, 5])

Evaluating the 7 sets containing closure/interior:

$A = (0, 1) \cup (1, 2) \cup \{3\} \cup (\mathbb{Q} \cap [4, 5])$
$c(A) = [0, 2] \cup \{3\} \cup [4, 5]$
$i(c(A)) = (0, 2) \cup (4, 5)$
$c(i(c(A))) = [0, 2] \cup [4, 5]$
$i(A) = (0, 1) \cup (1, 2)$
$c(i(A)) = [0, 2]$
$i(c(i(A))) = (0, 2)$

Notice that all 7 sets are mutually distinct:

$A \ne c(A)$ (e.g. $[4, 5] \setminus \mathbb{Q}$ is in $c(A)$ but not $A$).
$c(A) \ne i(c(A))$ ($3 \in c(A)$, not in $i(c(A))$).
$i(c(A)) \ne c(i(c(A)))$ (endpoints $0, 2, 4, 5$).
$c(i(c(A))) \ne c(i(A))$ ($[4, 5]$ is missing from $c(i(A))$).
$c(i(A)) \ne i(c(i(A)))$ ($0, 2 \in c(i(A))$).
$i(c(i(A))) \ne i(A)$ ($1 \in (0, 2)$ but $1 \notin i(A)$).

Taking the complements of these 7 sets yields another 7 distinct sets, completely disjoint from the first 7 because no set here is both open and closed! Thus, $A$ generates exactly 14 distinct sets.

Honors / Proof Challenge Example 2.3: Subbase Generation & Alexander's Subbase Characterization

Let $X$ be a set and let $\mathcal{S} \subseteq \mathcal{P}(X)$ be an arbitrary collection of subsets such that $\bigcup_{S \in \mathcal{S}} S = X$.

Prove rigorously that the collection $\mathcal{B}$ of all finite intersections of elements of $\mathcal{S}$:

\mathcal{B} = \left\{ \bigcap_{i=1}^k S_i : k \in \mathbb{N}, \; S_i \in \mathcal{S} \right\}

satisfies the Basis Criterion Theorem (Theorem 2.5), and therefore generates a unique topology $\mathcal{T}(\mathcal{S})$ on $X$.

Prove that $\mathcal{T}(\mathcal{S})$ is the coarsest (smallest) topology on $X$ that contains $\mathcal{S}$.

1. Proof of the Basis Criterion for Finite Intersections

Let $\mathcal{B} = \left\{ \bigcap_{i=1}^k S_i : k \ge 1, S_i \in \mathcal{S} \right\} \cup \{X\}$. We must verify the two conditions of Theorem 2.5:

Condition 1 (Covering): By hypothesis, $\bigcup_{S \in \mathcal{S}} S = X$. Since $\mathcal{S} \subseteq \mathcal{B}$, every $x \in X$ belongs to some $S \in \mathcal{S} \subseteq \mathcal{B}$. Thus $\bigcup_{B \in \mathcal{B}} B = X$.

Condition 2 (Intersection Property): Let $B_1, B_2 \in \mathcal{B}$. By definition of $\mathcal{B}$:

B_1 = \bigcap_{i=1}^m S_i \quad \text{and} \quad B_2 = \bigcap_{j=1}^n S'_j

for some $S_1, \dots, S_m, S'_1, \dots, S'_n \in \mathcal{S}$. Now compute their intersection:

B_1 \cap B_2 = \left( \bigcap_{i=1}^m S_i \right) \cap \left( \bigcap_{j=1}^n S'_j \right) = S_1 \cap \cdots \cap S_m \cap S'_1 \cap \cdots \cap S'_n

Notice that $B_1 \cap B_2$ is itself a finite intersection of $m + n$ elements of $\mathcal{S}$! Therefore:

B_3 = B_1 \cap B_2 \in \mathcal{B}

For any $x \in B_1 \cap B_2$, simply choosing $B_3 = B_1 \cap B_2 \in \mathcal{B}$ trivially satisfies:

x \in B_3 \subseteq B_1 \cap B_2

Condition 2 holds with equality! By the Basis Criterion (Theorem 2.5), $\mathcal{B}$ forms a valid basis for a topology on $X$.

2. Minimality of $\mathcal{T}(\mathcal{S})$

Let $\mathcal{T}'$ be any topology on $X$ containing $\mathcal{S}$ (i.e. $\mathcal{S} \subseteq \mathcal{T}'$).

Since $\mathcal{T}'$ is closed under finite intersections, every finite intersection of sets in $\mathcal{S}$ must belong to $\mathcal{T}'$. Thus $\mathcal{B} \subseteq \mathcal{T}'$.
Since $\mathcal{T}'$ is closed under arbitrary unions, every union of elements in $\mathcal{B}$ must belong to $\mathcal{T}'$.
By definition, $\mathcal{T}(\mathcal{S})$ consists precisely of all arbitrary unions of sets in $\mathcal{B}$.
Therefore:

\mathcal{T}(\mathcal{S}) \subseteq \mathcal{T}'

Hence, $\mathcal{T}(\mathcal{S})$ is contained in every topology containing $\mathcal{S}$, which proves that $\mathcal{T}(\mathcal{S})$ is the unique coarsest topology on $X$ containing $\mathcal{S}$. $\blacksquare$