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

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.

1. 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.

1. 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?

• yes, your interpretations are correct. – uniquesolution Mar 27 '19 at 12:42
• Excellent, thank you @uniquesolution :) – jjepsuomi Mar 27 '19 at 12:46

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.

• Thank you for your reply @HennoBrandsma. I don't know if it's the fact that my native language is not english but I did not get what you meant by this part: "...$B$ is compact in the subspace topology (from $X$) wrt the weak topology". Are you saying here that: "$B$ is compact and $B\in \tau_B =\{B\cap C\,|\,C\in\tau\}$, where $\tau$ is a weak topology on $X$"? – jjepsuomi Mar 27 '19 at 22:36
• @jjepsuomi almost: $\tau_B$ is what you described (the subspace topology wrt the weak topology on $X$) and then $(B, \tau_B)$ has to be compact. – Henno Brandsma Mar 27 '19 at 22:38
• Excellent ^^ to be 100% sure. I'm gonna verify it once more: $B$ is called weakly compact, if every $C\in \tau_B$ is compact. Is that it? :) – jjepsuomi Mar 27 '19 at 22:42
• @jjepsuomi No! just if $(B, \tau_B)$ is compact. – Henno Brandsma Mar 27 '19 at 22:43
• Okay, shame :/ I have to review what is the difference of a compact topological space and a compact set. Maybe I'm thinking too much in terms of sets, don't know. – jjepsuomi Mar 27 '19 at 22:44

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.

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 $$\mathbb R$$).

• Dear @LeeMosher thank you for your help. About the part 3, could you elaborate and put it in more simple terms? I'm not so proficient with the terminology yet. For instance, it is unclear what "compact with respect to the subspace topology on $B$ that is induced from $\tau$" means. What is a subspace topology? What does "$B$ that is induced from $\tau$" mean? – jjepsuomi Mar 27 '19 at 13:11
• I did some quick research and modified my 3rd interpretation based on how I understood your comment. Is this correct?: Let $\tau$ be a weak topology on a set $A$. The subset $B\subset A$ is weakly compact, if it is compact and $B\in\tau_{B}$, where $\tau_{B}=\{B\cap C\,|\,C\in\tau\}$. – jjepsuomi Mar 27 '19 at 13:25
• $\tau_B$ is indeed what I referred to as the subspace topology on $B$ that is induced from $\tau$. However, $B \in \tau_B$ is always true, regardless of whether $B$ is compact, because $B = B \cap A$ and $A \in \tau$. Think about $[0,1]$, which is a compact subset of $\mathbb R$, but is not an open subset of $\mathbb R$, i.e. it is not an element of the usual topology on $\mathbb R$. – Lee Mosher Mar 27 '19 at 13:44
• I added some comments regarding item 3. – Lee Mosher Mar 27 '19 at 13:50