# Weight of $[0,1]^k\le k$, where $k$ is an infinite cardinal

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?

• Is there a reason we care about the topology on $[0,1]?$ Mar 24, 2020 at 13:51
• I don't know this: I use the usual topology, since my text use it. Mar 24, 2020 at 13:54
• 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.
– user762847
Mar 24, 2020 at 18:19
• Unfortunately this is true: I forgot it; forgive my forgetfulness. Mar 24, 2020 at 18:21
• 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.
– user762847
Mar 24, 2020 at 18:24

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 $$\kappa$$.

• 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? Mar 25, 2020 at 11:19
• @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 $k$. Mar 25, 2020 at 11:57
• 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? Mar 25, 2020 at 12:40
• @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. Mar 25, 2020 at 13:23

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 $$k$$.

• 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 $[0,1]$. Mar 24, 2020 at 17:56
• 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 $w([0,1]^k)\le k$. Mar 24, 2020 at 17:57
• Is this what you want say? Mar 24, 2020 at 17:57
• Anyway I think I found an another proof: now I edit the question. Mar 24, 2020 at 17:58
• @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.
– user762847
Mar 24, 2020 at 18:04