Question concerning a topology induced by a particular metric.

[jmjlt88]

The question comes from the Munkres text, p. 133 #3.

Let X_n be a metric space with metric d_n, for n ε Z₊.

Part (a) defines a metric by the equation

ρ(x,y)=max{d₁(x,y),...,d_n(x,y)}.​

Then, the problem askes to show that ρ is a metric for the product space X₁ x ... x X_n.

When I originally started the problem, after showing that ρ is indeed a metric, I went on to try to show that the topology on X₁ x ... x X_n induced by ρ is the same as the product topology (or we can say the box topology since our product is finite and therefore the two topologies coincide) on X₁ x ... x X_n. I am fairly confident that I showed the ρ-topology is finer than the product toplogy, but I am having issues proving the converse. It boils down the following inequality;

d_i(x,y)≤ρ(x,y)≤____d_i(x,y).​

If I can fill in the blank with some upperbound, then I believe that for each i,

B_di(x_i; ε/something) is a subset of B_ρ(x;ε)​

.After re-reading the question, I believe all that Munkres wants me to do is show that ρ indeed defines a metric. But, I am curious about the topology ρ generates (Perhaps, it is finer and that's it). Thank you for the help! :)

 

[micromass]

There are some weird things about your post:

The question comes from the Munkres text, p. 133 #3.

Let X_n be a metric space with metric d_n, for n ε Z₊.

Part (a) defines a metric by the equation

ρ(x,y)=max{d₁(x,y),...,d_n(x,y)}.​

But $x,y\in X_1\times ...\times X_n$, right? So how does $d_1(x,y)$ even make sense?

Then, the problem askes to show that ρ is a metric for the product space X₁ x ... x X_n.

When I originally started the problem, after showing that ρ is indeed a metric, I went on to try to show that the topology on X₁ x ... x X_n induced by ρ is the same as the product topology (or we can say the box topology since our product is finite and therefore the two topologies coincide) on X₁ x ... x X_n. I am fairly confident that I showed the ρ-topology is finer than the product toplogy, but I am having issues proving the converse. It boils down the following inequality;

d_i(x,y)≤ρ(x,y)≤____d_i(x,y).​

Why would this inequality help (even it would even make sense)? The product topology is defined by open sets, not by a specific metric. It is true that two metric topologies are equal if the metrics satisfy some inequality. But here you don't know that the product topology comes from a metric, right??

 

[jmjlt88]

Micro, that is a HUGE typo on my part.

Define

ρ(x,y)=max{d₁(x₁,y₁), . . . , d_n(x_n,y_n)}.​

In fact, the book does not have x and y in boldface, which it should.

I had no trouble proving ρ is metric (which I now know is all the exercise wanted). Unknowning, I continued. I let ∏U_i be a basis for the product topology on ∏X_i, where U_i is open in X_i for each i. Then, I let x be a point in ∏U_i. Since U_i is open in X_i and contains x_i, we can find an ε_i-ball B_di(x_i;ε_i) centered at the i^th coordinate of x and contained in U_i. Let ε = min {ε₁,...,ε_n}. Then, I showed for each i, B_ρ(x_i; ε) is a subset of B_di(x_i;ε_i); which implies x is a point in B_ρ(x;ε) which is a subset of ∏U_i. Hence, the ρ-topology is finer.

Edit: I now realize this is nonsense because ρ is defined for the product space. Major brain meltdown, sorry! :)

 

[micromass]

It really is true that the topology of $\rho$ is the product topology, but you seem to be confused by the notations.

Try to prove this:

$$B_\rho((x_1,...,x_n), \varepsilon) = B_{d_1}(x_1,\varepsilon) \times ...\times B_{d_n}(x_n,\varepsilon)$$

 

[jmjlt88]

Wow, that is tremendously easy and illustrating! :)

Thank you so much!

To show two topologies are the same, I keep trying to use the method of showing one is finer than the other and vice versa. But, I now see how easy it can be to simply show that their bases are the same. I think the text (Munkres) did that once, yet for some reason I have abandoned that approach. Thanks again Micro; you're are always extremely helpful!

 

[micromass]

Wow, that is tremendously easy and illustrating! :)

Thank you so much!

To show two topologies are the same, I keep trying to use the method of showing one is finer than the other and vice versa. But, I now see how easy it can be to simply show that their bases are the same. I think the text (Munkres) did that once, yet for some reason I have abandoned that approach. Thanks again Micro; you're are always extremely helpful!

This is a special case of course. You won't always be in the situation that the bases are the same. The usual technique of showing that one is finer that the other (and vice versa) works more generally.