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Let $(Y,d)$ a metric space and $(X,\cal T)$ a topological space so that for any $A\in\cal T_d$ and for any compact $K\in\mathcal P(X)$ we put $$ S(K,A):=\{f\in Y^X:f[K]\subseteq A\} $$so that we define the collection $$ \mathcal S:=\{S(K,A):(K,A)\in\cal K\times T_d\} $$ where $\cal K$ is the collection of compact spaces of $X$ with respect $\cal T$. So we observe that the empty set is (trivially) compact and thus if $Y$ is open then any $f\in Y^X$ is an element of $S(\emptyset,Y)$ so that $\cal S$ is a subbase for a topology $\mathcal T_K$ which we call compact-open topology.

Now Munkres says that the compact open topology is finer that pointwise topology $\cal T_p$, which we remember is just the product topology: so if $f\in A$ with $A\in\cal T_p$ then there exists $x_1,\dots,x_n\in X$ with $n\in\omega$ and $\epsilon\in\Bbb R^+$ such that $$ f\in\bigcap_{i=1}^n\pi^{-1}_{x_i}\Big[B\big(f(x_i),\epsilon\big)\Big]\subseteq A $$ so that for any $i=1,\dots ,n$ let's we put $$ K_i:={x_i}\quad\text{and}\quad B_i:=B\big(f(x_i),\epsilon\big) $$ and thus let's we prove that $$ \tag{1}\label{1}f\in S_f\subseteq\bigcap_{i=1}^n\pi^{-1}_{x_i}\Big[B\big(f(x_i),\epsilon\big)\Big] $$ where we put $$ S_f:= \bigcap_{i=1}^n S(K_i,B_i) $$ So we observe that $f(x_i)$ lies in $B\big(f(x_i),\epsilon\big)$ for any $i=1,\dots,n$ so that for any $i=1,\dots,n$ the inclusion $$ f[K_i]\subseteq B_i $$ holds and so $f$ lies in $S_f$; however, if $g$ lies in $S_f$ then for any $i=1,\dots, n$ the inclusion $$ g[K_i]\subseteq B_i $$ holds but we know that $$ g[K_i]=\{g(x_i)\}=\{\pi_{x_i}(g)\} $$ so that $g$ lies in $\pi^{-1}_{x_i}\Big[B\big(f(x_i),\epsilon\big)\Big]$ or rather in $\bigcap_{i=1}^n\pi^{-1}_{x_i}\Big[B\big(f(x_i),\epsilon\big)\Big]$.

So \eqref{1} proves that any open set of $\cal T_p$ is union of open set of $\mathcal T_K$ so that we finally conclude that $$ \mathcal T_p\subseteq\mathcal T_K $$

Well, first of all I would like to know if I well proved that $\cal S$ is a subase and $\cal T_p$ is contained in $\mathcal T_K$ and then I would like if it is possible to prove these things in another more simple way: could someone help me, please?

Antonio Maria Di Mauro
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1 Answers1

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I'm a little unsure of what you're asking, but here's my best interpretation.

  1. Did you prove that $\mathcal{S}$ is a subbase? I'm not totally sure what you are asking here. The compact-open topology is generally defined to be the smallest topology containing the subsets you defined, so by definition it is a subbasis. Do you have a different definition of the compact-open topology that you are working with?

  2. Did you prove $\mathcal{T}_p \subseteq \mathcal{T}_K$? Outside of some typos, it looks good to me.

  3. Could your proof be improved? I think the following is a little faster, but a proof should be about understanding and not optimization.

Let $$\mathcal{B}_p := \{\pi_x^{-1}(B_\epsilon(y)) \ | \ x \in X, y \in Y, \epsilon > 0\}.$$ This is a subbasis for $\mathcal{T}_p$. Notice that $\pi_x^{-1}(B_\epsilon(y)) = S(\{x\}, B_\epsilon(y)),$ so $\mathcal{B}_p \subseteq \mathcal{S}$ and hence $\mathcal{T}_p \subseteq \mathcal{T}_K$.

User203940
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  • Okay, so to be more precisely you observed that the equality $$ \pi_x^{-1}(B_\epsilon(y))={f\in Y^X:\pi_x(f)\in B_\epsilon(y)}={f\in Y^X:f(x)\in B(\epsilon(y)}={f\in Y^x:f[x]\subseteq B_\epsilon(y)}=S\big({x},B_\epsilon(y)\big) $$ hods so that $\cal S$ so that $\cal B_p$ is contained in $S$ and thus any open set of $\cal T_p$ is and open set of $\mathcal T_K$, right? – Antonio Maria Di Mauro Dec 22 '22 at 08:35
  • Anyway you understood: I tried to prove that $\cal S$ is a subbase and thus I asked if I did it well. So did I well prove that $\cal S$ is a subbase? Indeed I know that a collection is a subbase of any topology if and only it it is a cover: is this true? – Antonio Maria Di Mauro Dec 22 '22 at 08:35
  • @AntonioMariaDiMauro Your last comment is not correct. Any collection of subsets of a set $X$ is a subbase for the topology on $X$ that it generates, by definition. It does not matter that the subsets cover $X$ or not. By definition, the topology that the subbase generates is the smallest topology containing the subbase, which always exists, namely as the intersection of all topologies containing the subbase. The article https://en.wikipedia.org/wiki/Subbase has more details about this topic. – PatrickR Dec 28 '22 at 20:58
  • @PatrickR Sorry, but I do not understand well what is my mistake: indeed, if $\cal S$ is a subbase for a topology $\cal T$ in a set $X$ then for any $x∈X$ there exist $S_1,…,S_n∈S$ with $n∈ω$ such that $$x∈S_1∩⋯∩S_n⊆⋃S$$ which proves that $\cal S$ cover $X$; conversely, if $\cal S$ cover $X$ then the collection of finite intersection of $\cal S$ is a base for a topology on $X$ because it is closed for finite intersection and because it covers $X$. So as you can see a collection $S$ is a subbase for a topology in $X$ if and only it covers $X$: are you again think I am wrong? – Antonio Maria Di Mauro Dec 29 '22 at 18:10
  • @AntonioMariaDiMauro A subbase for a topology on $X$ need not cover $X$. Take a look at the two (equivalent) definitions of subbase in wikipedia. Def 1 says that the subbase generates the topology. Def 2 says that the set of finite intersections of elements of the subbase form a base for the topology. These finite intersections include the intersection of the empty subfamily, which is the whole space $X$. But the subbase itself need not cover $X$. – PatrickR Dec 30 '22 at 03:34
  • @PatrickR Well, unfortunately (at least I know this) the intersection of empty set is not defined in ZF as here is proved. Anyway I reaffirm that if $\cal S$ is a subbase for $X$ equipped with a topology $\cal T$ then by the wiki definition $2$ and by opennes of $X$ for any $ x∈X$ there exist $S_1,…,S_n∈\cal S$ not empty (the intersection of the empty set is not defined!!!!) with $n∈ω$ such that $$x∈S_1∩⋯∩S_n⊆⋃S$$ and this clearly proves that $\cal S$ cover $X$ by arbitrariness of $x∈X$, right? – Antonio Maria Di Mauro Dec 30 '22 at 09:03
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    @AntonioMariaDiMauro See https://en.wikipedia.org/wiki/Intersection_(set_theory)#Nullary_intersection. It is true that the intersection of an empty family is not defined as a set in ZF, as it would have to be the whole universe, which is not a set. But in the context of a fixed enclosing set $X$ on which we want to define a topology, the intersection of the empty family of subsets of $X$ makes perfect sense, and it is $X$ (run through the definition of intersection of a family and you will see). Also notice the sentence with "nullary intersection convention" after Def 2 in wiki. – PatrickR Dec 30 '22 at 10:10
  • @PatrickR Okay, so substantially (for confirmation) a subbase could be not a cover putting informally that $$\bigcap\emptyset:=X$$ right? – Antonio Maria Di Mauro Dec 30 '22 at 10:15
  • Right, it need not be a cover. But there is nothing informal about the empty intersection of a family of subsets of a set. It's just a fact that is easy to prove: Suppose $\mathscr{A}$ is a collection of subsets of $X$. The set $\bigcap\mathscr{A}$ consists of those $x\in X$ that belong to all $A\in\mathscr{A}$. If $\mathscr{A}$ is empty, all $x\in X$ vacuously satisfy that, since there are no $A\in\mathscr{A}$. – PatrickR Dec 30 '22 at 10:22
  • @PatrickR Well, surely your arguments work (apparently) but unfortunately they do not agree with ZF formalism as explained in the above linked hartkp's answer and as in your comment above you implicitly "preannounced" so that with respect ZF I think it is better to put $$⋂∅:=X$$ justifying the above position with your last arguments. However I think this could be a bit tedious discussion: at this point it is clear why with respect your definition a subbase could be not a cover whereas with respect my definition it must be it. So thanks for the discussion: I learnt a lot! ;-) – Antonio Maria Di Mauro Dec 30 '22 at 10:49
  • @AntonioMariaDiMauro You are welcome. Also, if you don't like to make use of the intersection of the empty family, that's fine. In that case, note that in Def 2 in wikipedia they are careful to add the phrase "together with the set X" in the first sentence. That's needed so that Def 2 matches Def 1. Also note in the next sentence "every proper open set in T can be written ...", again hinting to the fact that if you don't start with a cover, you have to add X explicitly to the topology. – PatrickR Dec 30 '22 at 20:28