# Thread: Indiscrete topology and metric

1. ## Indiscrete topology and metric

I have a small trouble while trying to grasp which fact is described by the following statement:
"If a set X has two different elements, then the indiscrete topology on X is NOT of the form $\mathcal{T}_d$ for some metric d on X. Why? In particular, not every topology comes from a metric."

1. Now I tend to interpret the above statement that whenever X has (at least) two different elements and d is a metric on X, then $\mathcal{T}_d \neq \{\emptyset, X\}$. Then it makes sense to see a set with at least two elements as an example where the topology induced by a metric does not produce the empty set and the whole of X as the only open sets.

2. But the word some disturbs me. Because it allows for another interpretation, namely: whenever X has at least two elements there exists a metric d such that $\mathcal{T}_d \neq \{\emptyset, X\}$.

So before I head off to answer "why?" I would like to know: What do they mean exactly?

2. Originally Posted by HAL9000
I have a small trouble while trying to grasp which fact is described by the following statement:
"If a set X has two different elements, then the indiscrete topology on X is NOT of the form $\mathcal{T}_d$ for some metric d on X. Why? In particular, not every topology comes from a metric."?
Do you know what it means to say that a topological space is $\math{T}_2~?$
For each pair of distinct points there are two disjoint open sets each containing one of the points. Every metric space is $\math{T}_2$.

Is that possible in your space?

3. Yeah, I see. This makes it clearer. But they don't presuppose any knowledge of the hausdorff property, because it is defined later on. So the answer to the why?-question should be given on more or less "intuitive" grounds.

Anyway, $(X,\mathcal{T}_d)$ is hausdorff iff d is a metric, whereas X equipped with the indiscrete topology $\mathcal{I}$ is not hausdorff. This means that $\mathcal{T}_d$ is "richer" or in other words $\mathcal{I}\subsetneq\mathcal{T}_d$. With that hint it's all straightforward. But again they didn't presuppose it explicitly.

Now it seems that in the context of the original statement my first interpretation of "some" was correct. Here "some" in fact means "any" and points to an universal and not an existence argument.

Now let me try to answer the why?-question as if I had never heard of the hausdorff property. Assume that $X$ has at least two different elements. Supply $X$ with a metric $d$. We will show that in fact there are sets in $\mathcal{T}_d$ which are not in $\mathcal{I}$. Suppose that $\mathcal{I}=\mathcal{T}_d$. This means that X is the only open set in the metric space sense. By definition it contains a ball $B_\varepsilon(x)\subseteq X$ around each of its points $x\in X$. If $B_\varepsilon(x)\subsetneq X$ then there exists $y \in X$ different from x such that $y \notin B_\varepsilon (x)$. Define $r:=d(x,y)-\varepsilon$. Then $B_r(y)$ is a ball such that $B_r(y) \cap B_\varepsilon (x)=\emptyset$. But since $\mathcal{I}=\mathcal{T}_d$ we must have either $B_r(y)=\emptyset$ or $B_\varepsilon (x)=\emptyset$. Either case is impossible. This shows that in fact for every $x \in X$ we have the ball $B_\varepsilon (x)=X$. (OK, I see that I implicitly use the hausdorff property here.) But my question is: How should I interpret this last result? Does it now mean that whatever $\varepsilon>0$ we have $B_\varepsilon (x)=X$? This would allow to choose that epsilon arbitrarily small, so that I can infer that the distance $d(x,y)=0$ for $x \neq y$. Can I do this?

4. The above is just over the top.
There are many topological spaces that are not metrizable.
In metric space there must more than two open sets if there are at least two points.

5. OK, you're right. Thank you.