Clarifying topological concepts: induced/weak/initial topology and weakly compact

Question

I am a self-studying student on functional analysis and I have multiple times bumped into the same definitions from general topology that are not 100% clear for me. For this reason, I wanted to post these definitions to this post and ask the community if my understanding is correct or not.

Next, I will go through the definitions (taken from Wikipedia) with added interpretations of my own.

1. A topology $\tau$ induced on set $A$ by function $f$ is the coarsest topology such that $f$ is continuous.

My interpretation: $\tau$ is the smallest set of subsets of $A$, for which $f:\tau\to \text{codomain($f$)}$ is continuous.

2. The weak topology (or initial topology) $\tau$ on a set $A$, with respect to a family of functions $\mathcal{F}$ on $A$, is the coarsest topology on $A$ that makes those functions continuous.

My interpretation: $\tau$ is the smallest set of subsets of $A$, for which all the functions $f:\tau\to \text{codomain($f$)},\;f\in\mathcal{F}$ are continuous.

3. One may call subsets $B$ of a topological space weakly compact, if they are compact with respect to the weak topology.

My interpretation: Let $\tau$ be a weak topology on a set $A$. The subsets $B\subset A$ are called weakly compact, if they are compact and $B\in\tau$.

My question for the community is: Are my interpretations correct?

Answer 1

In this rather long answer I explain a part of the basic theory on initial topologies which are introduced on a set $X$ that has a set of maps $f_i: X \to Y_i, i \in I$ to topological spaces and where $X$ gets the minimal topology that makes all $f_i$ continuous (w.r.t. the topologies that the $Y_i$ already have). This construct is called the "weak topology" in analysis. This is basically the same as your interpretation 2.

In this post we see that we have a natural subbase for that weak topology, namely the set $\mathcal{S}= \{f_i^{-1}[U]: U \text{ open in } Y_i, i \in I\}$, and the Alexander subbase lemma tells us that $X$ is compact in the weak topology (or weak compact as it's called as well) iff every open cover of $X$ by members of $\mathcal{S}$ has a finite subcover.

Your final interpretation is nonsense: if $B \subseteq X$ is compact in the weak topology, this does not imply that $B$ is a member of that weak topology (i.e. open), in most practical cases this will not be the case. It just means that $B$ is compact in the subspace topology (from $X$) wrt the weak topology, and by the transitive law I showed in my post, we could also say that it is compact in the weak topology induced on $B$ wrt the restriction maps $f_i\restriction_B: B \to Y_i$ and a similar Alexander criterion of compactness applies.

Answer 2

Your interpretations for 1 and 2 are correct (except that $f : A \to \text{codomain}(f)$, not $f : \tau \to \text{codomain}(f)$). But they hide a very important part of the process, and from what you write I cannot tell whether you are aware of this.

A topology $\tau$ must satisfy some axioms: (1) the intersection of any two elements of $\tau$ is an element of $\tau$; (2) the union of any collection of elements of $\tau$ is an element of $\tau$; (3) the entire set, and the empty set, are elements of $\tau$.

Your version of 1, for example, says that $\tau$ is the set of all subsets of $A$ the form $f^{-1}(U)$ where $U$ is an element of the topology of $\text{codomain}(f)$.

This begs a big question: Does $\tau$ satisfy the axioms it is supposed to satisfy?

The answer is: Yes it does, but that requires some proof.

The outline of the proof is: Each of the three axioms (1), (2), and (3) for the topology on $\text{codomain}(f)$ implies the same axiom for $\tau$ itself.

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Your interpretation for 3 has an error, the correct statement would be that $B$ is weakly compact if it is compact with respect to the subspace topology on $B$ that is induced from $\tau$.

Added in response to a comment from the OP: I don't know how to express compactness of $B$ directly in terms of $\tau$ other than to repeat the standard definition: for every collection of elements $\{U_i\}_{i \in I}$ of $\tau$ such that $B \subset \cup_i U_i$, there exists a finite subset $\{i_1,...,i_N\} \subset I$ such that $B \subset \cup_{n=1}^N U_{i_n}$. This does not imply that $B \in \tau$. Think about the compact subset $[0,1] \subset \mathbb R$ which is not an open subset of $\mathbb R$ (i.e. is not an element of the usual topology on $\mathbb R$).