# Trying to understand “metric topology”

I have just started studying topology and I am having some difficulty in grasping the concept of metric topology.

As an example, let $(X,d)$ be a metric space and let $T_d$ = $\{G\subset X$|$G$ is an open set in $(X,d)\}$.

It is easy to verify that $T_d$ is a topology on $X$, what i don't quite follow is, that, why do we call it the topology induced by the metric $d$ ?

How is the metric $d$ inducing the topology $T_d$ ?

What is meant by saying that, the given topology is induced by a metric?

Also does this also mean that we can categorize topologies as the ones being induced by metrics and the ones that are not induced by metrics ? In that case would the indiscrete topology $\{\emptyset,X\}$ fall into the second category of topologies?

• Since the definition of $T_d$ depends on $d$? Or maybe $T_{(X,d)}$ is more adequate? – Lehs Feb 7 '15 at 11:35
• Without the metric $d$ how you define the concept of open set in the set $X$ ? – Mauro ALLEGRANZA Feb 7 '15 at 11:39

## 2 Answers

A topology on a set $X$ is a system ${\cal T}$ of distinguished subsets of $X$ which then are called open. When $X$ is large, say $={\Bbb R}$, then it is impossible to print out the list of the open sets one has in mind. Instead one sets up a test and calls a set $A\subset X$ open when it passes this test.

One possibility for such a test is defining a metric $d$ on $X$ and calling $A$ open if $\ldots$ (fill in the details). If $X$ is coordinatized somehow then the coordinates can be used for defining $d$, the simplest example being $d(x,y):=|x-y|$. Given $d$ no more data are needed to define a topology ${\cal T}_d$ on $X$ which is related to $d$ in the way you know. Therefore it is allowed to call this topology induced by $d$.

Not all topologies one would like to work with can be encaptured by a metric. ${\cal T}=\{\emptyset, X\}$ is an example, but not an interesting one. An important example is the following: Take as $X$ the space of all continuous functions $f:\>[0,1]\to{\mathbb R}$. You want to capture the idea that $$\lim_{n\to\infty} f_n=f \qquad\Longleftrightarrow\qquad \lim_{n\to\infty} f(x)=f(x) \quad \forall x\in[0,1]\ ,$$ called topology of pointwise convergence. This cannot be accomplished with a metric on $X$.

What this means is that the topology $T_d$ is generated by the "d-open" sets, i.e., the sets $G$ in the form {$x$ $\in$ $X$ $/$ $\exists$ $\delta$ $>$ $0$; $B(x,\delta)$ $\subset$ $G$}. Where, $B(x,\delta)$ $=$ {$a$ $\in$ $X$ $/$ $d(a,x)$ < $\delta$}.

In other words, when we discuss openness, we discuss it with respect to the above definition.

Example: in $(R, d_0 = usual-metric)$, $]0,1[$ is open and so $]0,1[$ $\in$ $T_{d_0}$; and $[1,2[$ is not open, so $[1,2[$ $\notin$ $T_{d_0}$.