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Let $P=\prod_{n=1}^{\infty} I_{n}$ with $I_{n}=[0,1]$

I must define two metrics on $P$ that generate two different topologies.

My idea:

$d_{1}:P\times P \rightarrow \mathbb{R}$

$d_{1}(x,y)=0$ if $x=y$

$d_{1}(x,y)=1$ if $x\neq y$

this is the discrete metric, in which all points are open and then, the respective topology is the discrete topology relative to $P$.

$d_{2}:=$ the usual metric on $\mathbb{R^n}$

which gives the usual topology relative to $P$ and is strictly coarser than the discrete topology.

Is my idea correct?

I'm worried about the fact that the product is infinite.

Thanks in advance.

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The metric on $\mathbb R^{n}$ does not define a metric on the infinite product. Define $d'((a_n),b_n)=\sup \{|a_n-b_n|: n \geq 1\}$. This is a metric and it is not equivalent to the discrete metric you have defined because $(\frac 1 n,\frac 1 n,\cdots)$ converges to $(0,0,...)$ in this metric but not in the discrete metric.

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As you say, the metric $d_2$ is defined on $\mathbb{R}^n$, so that alone won't suffice to equip $P$ with a metric. Given a family of metric spaces $(X_n,d_n)$, the function

$$ d(x,y) = \sum_{k \geq 1}\frac{\min\{d_k(x_k,y_k),1\}}{2^k} $$

is a metric on $\prod_n X_n$, and if I recall correctly it is topologically equivalent to the product topology with respect to the metric induced topologies on each $X_i$. In your case, each $X_n$ is $[0,1]$ with the usual distance.

In any case, to conclude we should see that $d_1$ and $d$ are not topologically equivalent, and for that it suffices to see that a point in $\prod_n X_n$ is not open for $d$ (but it is for $d_1$). Pick any $p \in \prod_n [0,1]$. Now, to see that $\{p\}$ is not open, let's show that for any $\varepsilon > 0$, $B_\varepsilon(p) \neq \{p\}$. In effect, since $p_1 \in [0,1]$ which has no isolated points, there exists $q_1 \in [0,1]$ with $|p_1 - q_1| < \varepsilon$. Thus, the point $(q_i)_{i \geq 1}$ with

$$ q_i = \cases{q_1 \text{ if $i = 1$} \\ p_i \text{ otherwise}} $$

is an element of $B_\varepsilon(p)$.

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The product metric on $\prod_n I_n$ is given by

$$d(x,y) = \sum_n \frac{1}{2^n} | x_n -y_n|$$

and this induces the product topology (cf. this answer), which makes the product space compact, while

$$\rho(x,y) = \sup_n |x_n - y_n|$$

also introduces a metric on this set where the set of all $e_n = (0,0,\ldots, 1, 0,\ldots)$ (the $1$ on place $n$) is a countable closed and discrete subspace, showing that the product is then very non-compact.

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