$[0,1]$ is compact. Why is $f:\{0,1\}^{\mathbb{N}}\ni (a_n) \mapsto \sum_{n=1}^\infty \frac{a_n}{2^n}\in\mathbb{R}$ continuous? Takeshi Saito's proof.

Question

I am reading "How to Learn Mathematics (New Edition)" (in Japanese) edited by Kunihiko Kodaira.

Few years ago, I posted my answer for this question.

My answer was copy and paste from the above book and I don't perfectly understand my answer.

$[0, 1]$ is compact.

Takeshi Saito's proof:
The discrete space $\{0, 1\}$ is finite. So $\{0, 1\}$ is compact. So, $\{0,1\}^{\mathbb{N}}$ is compact.
Let $f : \{0,1\}^{\mathbb{N}} \to \mathbb{R}$ be a function such that $f((a_n)) := \sum_{n=1}^\infty \frac{a_n}{2^n}$.
Then, $f$ is continuous, so $f(\{0,1\}^{\mathbb{N}}) = [0,1]$ is compact.

I filled gaps of the above proof as follows:
Since $\{0,1\}$ is finite and finite topological spaces are compact, $\{0,1\}$ is compact.
By Tychonoff's Theorem, $\{0,1\}^{\mathbb{N}}$ is compact.
And I don't undrstand why $f : \{0,1\}^{\mathbb{N}} \to \mathbb{R}$ such that $f((a_n)) := \sum_{n=1}^\infty \frac{a_n}{2^n}$ is continuous.
Since continuous image of a compact set is compact by a famous proposition, $f(\{0,1\}^{\mathbb{N}})$ is compact.
And $f(\{0,1\}^{\mathbb{N}})=[0,1]$.
So, $[0,1]$ is compact.

Please prove that $f : \{0,1\}^{\mathbb{N}} \to \mathbb{R}$ such that $f((a_n)) := \sum_{n=1}^\infty \frac{a_n}{2^n}$ is continuous.

Answer 1

We have $\displaystyle f=\sum_n\frac{\pi_n}{2^n}$, where $\pi_n\colon 2^{\Bbb N}\kern-5mu\longrightarrow \{0,1\}\kern5mu$ is the $n^{\scriptscriptstyle \rm th}$ projection.

These projections are continuous (by definition of the product topology), hence $f$ —being a normally convergent series of continuous functions— is continuous.