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!
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 $\mathcal T$.