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Statement

Let be $k$ an infinite cardinal and let be $[0,1]$ equipped with the usual topology. Well I ask if the weight of $[0,1]^k$ is such that $\le k$.

Proof. Previously as reference I say that the weight $w(X)$ of a topological space $X$ is the following quantity:

$$ w(X)=\min\{|\mathcal{B}|:\mathcal{B}\text{ is a base for } X \} + \aleph_0 $$ that obviously is such that $\ge\aleph_0$. Moreover we remember that a topological space $X$ is second countable iff there exist a countable basis $\mathcal{B}$ for it and so we remember that the second countability property is hereditable on subspace and on product -if each factor of product have this property.

So now we prove the satement. First of all we remember that $\Bbb{R}$ is second countable and so for what we above observed it results that $[0,1]^k$ is second countable and so there exist a countable basis for it and so $w([0,1]^k)=\aleph_0\le k$ since $\aleph_0$ is the first infinite cardinal.

So is my proof correct? Could someone help me?

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  • $\begingroup$ Is there a reason we care about the topology on $[0,1]?$ $\endgroup$ Mar 24, 2020 at 13:51
  • $\begingroup$ I don't know this: I use the usual topology, since my text use it. $\endgroup$ Mar 24, 2020 at 13:54
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    $\begingroup$ You are claiming, without proof, that because $\mathbb{R}$ is second countable, then $[0,1]$ is also second countable. OK, fine. Now, $[0,1]^k$ is not second countable when $k$ is not countable. $\endgroup$
    – user762847
    Mar 24, 2020 at 18:19
  • $\begingroup$ Unfortunately this is true: I forgot it; forgive my forgetfulness. $\endgroup$ Mar 24, 2020 at 18:21
  • $\begingroup$ The proof that countable product of second countables is second countable, rather then the result is what is important to remember/understand here. In the proof, you only need to replace the cardinality of the exponent by $k$ and do the rest of the cardinality computation with it. Take into account that whenever you multiply $\aleph_0$ or add it to an infinite cardinal $k$, you get $k$ as result. $\endgroup$
    – user762847
    Mar 24, 2020 at 18:24

2 Answers 2

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The crux of the matter is that we have a countable base $\mathcal{B}$ for $[0,1]$ (say all rational intervals $(q,r)$ and sets $[0,q)$, $(q,1]$ for $q (< r) \in \Bbb Q$) and all standard basic elements depend only on finitely many coordinates.

So form the following base for $[0,1]^\kappa$, where $\kappa$ is an infinite cardinal number:

$$\mathscr{B}= \Biggl\{\bigcap_{i \in F} \pi_\alpha^{-1}[B_\alpha] \mid F \subseteq \kappa \text{ finite }, \forall \alpha \in F: B_\alpha \in \mathcal{B} \Biggl\}$$

It's easy to see it is a base for the open sets and the only minor issue is computing its size: For each $n \in \omega$, we have $\kappa^n = \kappa$ many choices for $F$ of size $n$ and for each of these $\kappa$ choices we have $\aleph_0^n = \aleph_0$ many choices of base elements $B_\alpha$. So for each fixed $n$ we have $\kappa \cdot \aleph_0 = \kappa$ many options. Now we can let $n$ vary and we have a total of $\aleph_0 \cdot \kappa$ many options and this again just equals $\kappa$ by standard cardinal arithmetic. So $w([0,1]^\kappa) \le \kappa$ as the existence of this base shows.

Now, any base $\mathscr{B}'$ for $[0,1]^\kappa$ must have at least $\kappa$ members, for suppose it had $\aleph_0 \le \lambda < \kappa$ members, then the canonical base for $[0,1]^\kappa$ (all product open sets that depend on finitely many coordinates) would have a subcollection $\mathscr{C}$ of size $\lambda$ that was also a base (a fact I call the "thinning out lemma", it's thm. 1.1.15 in Engelking's classic General Topology) and now for each $C \in \mathscr{C}$ we have a finite set $s(C)$ of coordinates on which the set $C$ is not the whole space (its support), and $|\bigcup \{s(C): C \in \mathscr{C}\}| \le \aleph_0 \cdot \lambda = \lambda$. So some $\alpha \in \kappa$ exists that is not in that union and then $\pi_\alpha^{-1}[[0,\frac12)]$ is open but does not contain a set from $\mathcal{C}$, contradiction.

So the weight of $[0,1]^\kappa$ is exactly $\kappa$.

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  • $\begingroup$ Your proof it is clear to me. However I'd like to discuss more accurately the inequality $w([0,1]^k)\le k$, since the definition of weight that I use is not conventional. Well if I understood the proof we proved that $|\mathcal{B}|=k$ and so, since $\mathcal{B}$ is a basis, it results that $\text{min}\{|\mathcal{D}|:\mathcal{D}\text{ is a base for } [0,1]^k\}\le|\mathcal{B}|$ and so finally $w([0,1]^k):=\text{min}\{|\mathcal{B}|:\mathcal{B}\text{ is a base for } [0,1]^k \}+\aleph_0\le|\mathcal{B}|+\aleph_0=k+\aleph_0=k$; right? $\endgroup$ Mar 25, 2020 at 11:19
  • $\begingroup$ @AntonioMariaDiMauro Your definition of weight is completely standard. It’s the minimal size of a base for $X$. So because we already have one base of size $k$ the weight is $\le k$. The plus $=\aleph_0$ is just to ensure that the weight is always infinite, i.e. at least $\aleph_0$, so it has no effect on the upper bound $k$. $\endgroup$ Mar 25, 2020 at 11:57
  • $\begingroup$ Okay; so if we have a base of size $k$ and if we want add $\aleph_0$ to the definition of weight then it results that $w([0,1]^k):=\text{min}\{|\mathcal{B}|:\mathcal{B}\text{ is a base for } [0,1]^k \}\le|\mathcal{B}|=k$ and so adding $\aleph_0$ we have $w([0,1]^k)+\aleph_0\le k+\aleph_0=k$ that is the definition I above given, right? or rather is correct what I observed? $\endgroup$ Mar 25, 2020 at 12:40
  • $\begingroup$ @AntonioMariaDiMauro Rather $w([0,1]^k) = \min\{|\mathcal{B}|: \mathcal{B} \text{ is a base for } [0,1]^k\} +\aleph_0 \le |\mathcal{B}| + \aleph_0 = k+ \aleph_0 = k$, but that's a bit formal and unneeded as all cardinals here are infinite anyway. $\endgroup$ Mar 25, 2020 at 13:23
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You can take as basis of the topology the sets of the form $\times_{i\in k}U_i$, where the $U_i\subset[0,1]$ are open and only finitely many of them are different from $[0,1]$. We can take those factors that are different from $[0,1]$ to only be open intervals with rational end-points (here calling open also $[0,r)$ and $(r,1]$).

There are $k$ choices of which $n\in\mathbb{N}$ finitely many of the factors are different from $[0,1]$ and for each of them there are $\aleph_0^n=\aleph_0$ possible open intervals with rational end-points to take. This gives $k\times\aleph_0=k$ sets in which $n$ of the factors are not $[0,1]$, for each $n\in\mathbb{N}$. Adding together the count for each $n=0,1,2,3,...$, it gives in total $\aleph_0\times k=k$ sets in this basis.


Yes, the topology of $[0,1]$ is relevant. For example, if $[0,1]$ had the discrete topology and $k$ were smaller than the cardinality of $[0,1]$, then the weight of $[0,1]^k$ would be equal to the cardinality of $[0,1]$, which is larger than $k$.

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  • $\begingroup$ Sorry, it seems that your proof is not very clear to me: so to summarize now I rewrite it. I rewrite your proof: unfortunately it seems to me not very clear. So let be $k$ an infinite cardinal and $X_j=[0,1]$ for any $j\in k$; so we consider $X:=[0,1]^k$ and we know that if $\mathcal{B}_j$ is a basis of $[0,1]$ for any $j\in k$ then the collection $\mathcal{B}=\{\pi^{-1}_{j_1}(B_1)\cap...\cap\pi^{-1}_{j_n}(B_n):B_i\in\mathcal{B}_{j_i}\land j_i\in k\land i=1,...,n\land n\in\Bbb{N}\}$ is a basis for $[0,1]$. $\endgroup$ Mar 24, 2020 at 17:56
  • $\begingroup$ Since if for any $x,y\in\Bbb{R}$ there exist $r\in\Bbb{Q}$ such that $x<r<y$ any basis we can claim that the set $\{(r,s)\cap[0,1]:r,s\in\Bbb{Q}\}$ is a basis for $[0,1]$ that additionally is numerable and so the collection $\mathcal{Q}=\{\pi^{-1}_{j_1}((r_1,s_1)\cap[0,1])\cap...\cap\pi^{-1}_{j_n}((r_n,s_n)\cap[0,1]):r_i,s_i\in\Bbb{Q}\land j_i\in k\land i=1,...,n\land n\in\Bbb{N}\}$ is a basis of $[0,1]^k$ such that $|\mathcal{Q}|=k$ and so $w([0,1]^k)\le k$. $\endgroup$ Mar 24, 2020 at 17:57
  • $\begingroup$ Is this what you want say? $\endgroup$ Mar 24, 2020 at 17:57
  • $\begingroup$ Anyway I think I found an another proof: now I edit the question. $\endgroup$ Mar 24, 2020 at 17:58
  • $\begingroup$ @AntonioMariaDiMauro Yes, that's all. You have a countable basis of $[0,1]$. Then with it you get the usual basis with which one defines/constructs the product topology. Computing cardinality you get that that basis has $k$ elements. $\endgroup$
    – user762847
    Mar 24, 2020 at 18:04

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