KindiakMath

A Taste of Topology

Definition 1. Let K be any set. A collection \mathcal T \subseteq \mathcal P(K) is a topology on K if it satisfies the following properties:

  • \emptyset, K \in \mathcal T,
  • For U_\alpha \in \mathcal T, \displaystyle \bigcup_\alpha U_\alpha \in \mathcal T.
  • For U_1,\dots,U_n \in \mathcal T, \displaystyle \bigcap_{i=1}^n U_i \in \mathcal T.

For the last property, it suffices to verify the result for n =2. If such a topology exists, we call (K,\mathcal T) a topological space. Elements of T are called open sets.

Example 1. For any set K, (K, \{\emptyset, K\}) is a topology known as the trivial topology, and that (K, \mathcal P(K)) is a topology known as the discrete topology.

Problem 1. Let K be a set. The collection \mathcal B of subsets of K forms a topological basis for K if it satisfies the following properties:

  • For any x \in K, there exists B \in \mathcal B such that x \in B.
  • For any x \in K and any B_1,B_2 \in \mathcal B, if x \in B_1 \cap B_2, then there exists B_3 \in \mathcal B such that x \in B_3 \subseteq B_1 \cap B_2.

Prove that the collection \mathcal T of subsets defined by

\mathcal T := \{U \in \mathcal P(K):(\forall x \in U\ \exists B \in \mathcal B: x \in B \subseteq L)\}

is topology on K, known as the topology generated by \mathcal B.

(Click for Solution)

Solution. We prove the last property as it is the most interesting. If K_1 \cap K_2 = \emptyset then we are done.

Suppose otherwise and fix x \in U_1 \cap U_2. For each i, find B_i \in U_i such that x \in B_i \subseteq U_i. Then x \in B_1 \cap B_2 \subseteq U_1 \cap U_2. Find B_3 \subseteq B_1 \cap B_2 that contains x, so that x \in B_3 \subseteq B_1 \cap B_2 \subseteq U_1 \cap U_2. Hence, U_1 \cap U_2 \in \mathcal T, as required.

Example 2. Let (K, \mathcal T_K) and (L, \mathcal T_L) be topological spaces and let \mathcal B_K be a basis for K, \mathcal B_L be a basis for L. Define the box topology \mathcal T \subseteq \mathcal P(K \times L) as the topology generated by the basis \mathcal B_K \times \mathcal B_L.

Problem 2. Let (K, \mathcal T) be a topological space. For any L \subseteq K, define the collection \mathcal T_L \subseteq \mathcal P(K) of subsets by

\mathcal T_L := \{L \cap U : U \in \mathcal T\}.

Prove that (L, \mathcal T_L) is a topological space, known as a subspace topology.

(Click for Solution)

Solution. For the first property, \emptyset = L \cap \emptyset \in \mathcal L and L = L \cap K \in \mathcal L.

For the second property, for any \alpha, fix L \cap U_\alpha \in \mathcal T_L. Then by distributivity,

\displaystyle \bigcup_\alpha (L \cap U_\alpha) = L \cap \bigcup_\alpha U_\alpha = L \cap U \in \mathcal T_L,

since U := \bigcup_\alpha U_\alpha \in \mathcal T.

For the third property, for U_1,U_2 \in \mathcal T,

(L \cap U_1) \cap (L \cap U_2) = L \cap (U_1 \cap U_2) \in \mathcal T_L

since U_1 \cap U_2 \in \mathcal T.

Problem 3. Let (K, \mathcal T) be a topological space. Recall that elements of \mathcal T are called open. A set C \in \mathcal P(K) is called closed if K \backslash C is open (i.e. K \backslash C \in \mathcal T). Prove the following properties:

  • \emptyset, K are closed,
  • For closed sets C_\alpha \in \mathcal T, \displaystyle \bigcap_\alpha C_\alpha is closed.
  • For closed sets C_1,\dots,C_n \in \mathcal T, \displaystyle \bigcup_{i=1}^n C_i is closed.
(Click for Solution)

Solution. The results all follow by complementation, since

\displaystyle K\backslash \emptyset = K \in \mathcal T, \quad K \backslash K = \emptyset \in \mathcal T

and

\displaystyle K \backslash \bigcap_\alpha C_\alpha = \bigcup_\alpha (K \backslash C_\alpha) \in \mathcal T,\quad K \backslash \bigcup_{i=1}^n C_i =\bigcup_{i=1}^n (K \backslash C_i) \in \mathcal T.

—Joel Kindiak, 7 Feb 25, 2306H

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