# Metrics that define different topologies

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.

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.

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 $$B_\varepsilon(p)$$.

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.