Cartesian product indexed by a closed interval

Question

Let $I=[0,1]$ and $$X = \prod_{i \in I}^{} \mathbb{R}$$ That is, an element of $X$ is a function $f:I→\mathbb{R}$.

Prove that a sequence $\{f_n\}_n ⊆ X$ of real functions converges to some $f ∈ X$ in the product topology on $X$, if and only if it converges pointwise, i.e. for every $x ∈ I$, $f_n(x) → f(x)$ in the usual sense of convergence of sequences.

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I don't understand how a product indexed by an uncountably infinite set is like. I'm guessing a single element $f$ of $X$ is an uncountable set $(y_i)_{i \in I}$ in itself. But then how is $f(x)$ different from any other $f$? Is it the cartesian product indexed by $[0,x]$?

Note that I included the "prove" question just to show the context, but it isn't what I find problematic. It's probably easy enough if I get how the product works.

Answer 1

An element of $X = \prod_{i \in [0,1]} \mathbb{R}$ is just a function from $[0,1]$ to $\mathbb{R}$, i.e. a "choice" of a point in $\mathbb{R}$ for each $x \in [0,1]$. It's the same whether we write $(x_i)_{i \in [0,1]}$ or just the function $f$ that sends $i$ to $x_i$. Just writing $f$ is often easier in notations.

Likewise it's the same whether I write $(x,y,z) \in \mathbb{R}^3$ or the function $f: \{0,1,2\} \to \mathbb{R}$ defined by $f(0) = x, f(1) = y ,f(2) = z$.

Answer 2

Every single element of $X$ is simply a function: $I\to \mathbb{R}$. They differ as normal functions differ.

Answer 3

To make things easier to understand, let $I = {0, 1, 2}$. To say $$X = \prod_{i \in I} \mathbb{R}$$ means that an element of $X$ is an technically an element of $\mathbb{R^3}$. But this can be interpreted as a function from $I \to \mathbb{R}$, in the sense that

$$(a, b, c) \cong f \text{ such that} f(0) = a, f(1) = b, f(2)= c$$

To be accurate the function $f$ would be $\{(0, a),(1, b), (2, c)\}$. So, you could say

$$X = \prod_{i \in I} (\{i\} \times \mathbb{R})$$

where we consider the unordered Cartesian product. Now in your case, $I = [0,1]$, but the concept is the same.