0

The cantor space is defined as $\{0,1\}^{\mathbb{N}}$. Let $a=(a_k,k \geq 0)$ be a sequence in $(0, \infty)$ with $\sum_k a_k < \infty$. Let us define the metric $d_a$ as $$d_a ((x_k)_{k \in \mathbb{N}}, (y_k)_{k \in \mathbb{N}}) = \sum_{k \in \mathbb{N}} a_k |x_k - y_k|$$ in $\{0,1\}^{\mathbb{N}}$. Then a sequence $Z^n=((z_k^n))$ in $\{0,1\}^{\mathbb{N}}$ converges to $Z=(z_k)$ if and only if for all $k \geq 0$, $\lim_{n \rightarrow \infty} z_k^n=z_k$.

The question now is: Why is the topology induced by this metric $d_a$ (call it $\mathcal{T_d}$) the same as the product topology (call it $\mathcal{T}$), where the product topology is the smallest topology such that the coordinate maps are continuous? How can I prove that $\mathcal{T_d}=\mathcal{T}$?

Thanks in advance!

vaoy
  • 1,258
  • One way would be to prove that $\mathcal{T_d}$ satisfies the universal property, i.e. if $Y$ is any topological space and $f_i:Y\to {0, 1}$ is a sequence of continous functions then there exists a unique function $f:Y\to {0, 1}^{\mathbb{N}}$ such that $\pi_i\circ f=f_i$, where $\pi_i:{0,1}^{\mathbb{N}}\to{0,1}$ is the projection onto $i$-th coordinate (a.k.a. the coordinate map). If $\mathcal{T_d}$ satsfies this then it has to be the product topology. Of course the construction and uniqueness of $f$ is trivial. What is not obvious is that it is continous in $\mathcal{T_d}$. – freakish Jun 30 '17 at 09:52
  • Look at this question:https://math.stackexchange.com/q/361778/4280 and my detailed answer, which might give you ideas. – Henno Brandsma Jul 01 '17 at 22:16

1 Answers1

0

The statement about the sequence is actually:$$Z^n\to Z\text{ in }\{0,1\}^n\iff\pi_i(Z^n)\to\pi_i(Z)\text{ for all }i\in\mathbb N$$where $\pi_i:\{0,1\}^n\to\{0,1\}$ denotes the projection.

Then $\implies$ tells us that every projection is sequentially continuous w.r.t. $\mathcal T_d$.

Space $\langle\{0,1\}^{\mathbb N},\mathcal T_d\rangle$ is a metric space and functions that are sequentially continuous and have a metric space as domain are also continuous. By definition $\mathcal T$ is the coarsest topology that makes the projections continuous, so that $\mathcal T\subseteq\mathcal T_d$.

Conversely from $\impliedby$ we conclude that every sequence in $\{0,1\}^{\mathbb N}$ that converges wrt $\mathcal T$ will also converge wrt $\mathcal T_d$. That implies that $\mathcal T_d\subseteq\mathcal T$.

To make this more clear: let $z$ be a limitpoint of set $A$ wrt topology $\mathcal T$. Then a sequence $(z_n)$ can be constructed with $z_n\in A$, $z_n\neq z$ and $z_n\to z$ wrt $\mathcal T$. But then also $z_n\to z$ wrt $\mathcal T_d$ so $z$ is also a limitpoint of $A$ wrt $\mathcal T_d$. So if $A$ is closed wrt $\mathcal T_d$ (i.e. contains all its limitpoints) then it will also be closed wrt $\mathcal T$.

drhab
  • 151,093
  • The $\implies$ tells us that all projections are sequentially continuous. So you need to appeal to the fact that $\mathcal{T}_d$ is sequential (being metric) to see that they're continuous. By minimality of the product topology $\mathcal{T} \subseteq \mathcal{T}_d$. The reverse follows because the countable product of sequential topologies is sequential. – Henno Brandsma Jul 02 '17 at 13:33