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?

  • $\begingroup$ Since the definition of $T_d$ depends on $d$? Or maybe $T_{(X,d)}$ is more adequate? $\endgroup$ – Lehs Feb 7 '15 at 11:35
  • 2
    $\begingroup$ Without the metric $d$ how you define the concept of open set in the set $X$ ? $\endgroup$ – Mauro ALLEGRANZA Feb 7 '15 at 11:39

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 $[1,2[$ $\notin$ $T_{d_0}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.