Completely Linear

Let $X$ be any nonempty set. By $\mathcal P(X)$ we will denote the power set of $X$ , i.e. the set containing all possible subsets of $X$ :

$\mathcal P(X) = \{ A: A \subset X\}.$

A topology on $X$ is a subset $\mathcal T \subset \mathcal P(X)$ satisfying:

$X \in \mathcal T$ and $\emptyset \in \mathcal T$ ,
If $\mathcal A \subset \mathcal T$ , then $\bigcup_{A \in \mathcal A} A \in \mathcal T$ ,
If $\mathcal A \subset \mathcal T$ consists of finitely many elements, then $\bigcap_{A \in \mathcal A} A \in \mathcal T$ .

As usual, a function $f: (X, \mathcal T) \to (Y, \mathcal S)$ , where $(X, \mathcal T)$ and $(Y, \mathcal S)$ are topological spaces, is continuous if $f^{-1}(O) \in \mathcal T$ for any $O \in \mathcal S$ . It is easy to check that the composition of continuous functions is continuous.

Lemma: If $X$ is a non-empty set and $\{ \mathcal T_i: i \in I\}$ a family of topologies on $X$ (where $I$ is an arbitrary index set), then

$\mathcal T := \bigcap_{i\in I} \mathcal T_i$

is also a topology on $X$ .

Proof: It is straightforward to verify that $\mathcal T$ satisfies the axioms of a topology. ∎

Lemma: If $X$ is a non-empty set, $\{ (Y_i, \mathcal S_i: i \in I\}$ a family of topological spaces (where $I$ is an arbitrary index set) and for each $i \in I$ $f_i$ is a function from $X$ to $Y_i$ . Then, there is a minimal (i.e. smallest) topology $\mathcal T$ on $X$ such that each function $f_i$ is continuous as a function from $(X, \mathcal T)$ to $(Y, \mathcal S_i)$ .

Proof: Clearly, there exists a topology making each $f_i$ continuous (take the trivial topology $\mathcal P(X)$ ). Hence, we can consider

$\mathcal T = \bigcap_{i \in I} \big \tau,$

where the second intersection is taken over all topologies $\tau$ on $X$ making $f_i$ continuous. ∎

The topology in the previous lemma, being defined by $X$ , the topological spaces $(Y_i, \mathcal S_i)$ and the functions $f_i$ ( $i \in I$ ) is called the initial topology.

Given the family $(X_i, \mathcal T_i)$ of topological spaces, we can consider the cartesian product

$X := \prod_{i \in I} X_i := \{ x: I \to \sqcup_i X_i: x(i) \in X_i \text{ for every } i \in I\}.$

Now, the Axiom of choice yields that the product set $X$ is non-empty. In particular, we can consider the initial topology making all the projections

$\pi_i: X \to X_i, \quad \pi_i(x) = x(i)$

continuous. This topology is called the product topology on $X$ . We have the following important facts about product topologies:

Lemma: Let $(X_i, \mathcal T_i)$ be a collection of topological spaces, $(X, \mathcal T)$ and their product space with product topology. Then, the projection maps $\pi_i: X \to X_i$ are open, i.e. for $O \in \mathcal T$ we have $\pi_i(O) \in \mathcal T_i$ .

Proof: Let $O \in \mathcal T$ . ∎

Lemma: Let $(X_i, \mathcal T_i)$ be a collection of topological spaces, $(X, \mathcal T)$ their product space with product topology and $f: (Y, \mathcal S) \to (X, \mathcal T)$ a function. Then, $f$ is continuous if and only if $\pi_i \circ f$ is continuous for every $i \in I$ . Here, $\pi_i$ is again the projection $\pi_i: X \to X_i$ .

Proof: Contininuity of $f$ implies continuity of $\pi_i \circ f$ for any $i$ , since the projections $\pi_i$ are continuous by construction of the product topology.
Now, assume that $\pi_i \circ f$ is continuous for each $i$ . Given some $O \in \mathcal T$ , we need to show that $f^{-1}(O) \in \mathcal S$ . ∎

Completely Linear

Vector spaces: The algebraic essentials

Some Basics from Topology

Recent Posts

Recent Comments

Archives

Categories

Meta