Let $(X,d_1)$ and $(X,d_2)$ be two metric spaces which have the same infinite set $X$, but the different metrics $d_1$ and $d_2$.

Denote the collection of subsets $X$ by $S$, and the collection of all open subsets of $(X,d)$ by $U_{d}$.

Then, is it possible to know $|U_{d_1}|$ and $|U_{d_2}|$, the cardinalities of the collections of all open subsets of $(X,d_1)$ and $(X,d_2)$?

For example, if $X=\mathbb{R}$ and if the metric $d$ on $\mathbb{R}$ is defined to be the discrete metric, then $|U_d|=|S|.$

But, if $X=\mathbb{R}$ and if the distance function $d$ of $X$ is defined to be the usual Euclidean distance function, $|U_d|=c$.

In conculusion, what I'm asking is that is it possible to characterize $|U_d|$ of a metric space $(X,d)$, with $X$ being infinite, in terms of metric function $d$?

Also, is it possible to relate $|S|$ with $|U_d|$ of a metric space $(X,d)$ in terms of metric function $d$?

  • $\begingroup$ At a minimum you would need to be able to separate out the maximal $Y \subset X$ whose subspace topology is discrete, because if $|Y|=|X|$ then the cardinality of the topology of $X$ is $2^{|X|}$. If $|Y|<|X|$, then I am not sure what must necessarily happen. In particular I am not sure whether the statement depends on the continuum hypothesis. $\endgroup$
    – Ian
    Aug 16 '14 at 23:36
  • 1
    $\begingroup$ @Ian Why do you consider the subspace $Y$ and the subspace topology on $Y$? We just need to consider the topology on $X$. If the given topology on $X$ is the discrete topology which can be induced by the discrete metric, then the cardinality of the topology is $2^{|X|}$, which is same as the cardinality of the collection of all subsets of $X$. Then, the question is that if the given topology on $X$ is not the discrete one, can you obtain the corresponding metric $d$ on $X$ such that the given topology on $X$ is equal to the topology induced by the metric $d$? $\endgroup$
    – User
    Aug 17 '14 at 1:16
  • 1
    $\begingroup$ If the topology on all of $X$ is discrete then you're done, but I can make a disconnected union of a discrete space and a non-discrete space. If the former has the same cardinality as all of $X$ then you still get that the topology has cardinality $2^{|X|}$ even though $X$ itself is not discrete. It should be possible to metrize this idea as well. For example you could take a discrete space $Y$ (i.e. $d(x,y) = 1$ if $x \neq y$ and $x,y \in Y$), then have $Z$ homeomorphic to $[0,1]$, and assemble a metric space from their union with $d(y,z) = 1$ if $y \in Y$ and $z \in Z$. $\endgroup$
    – Ian
    Aug 17 '14 at 1:27

Let $X$ be an infinite set.

Note that a topology $\tau$ on X is a collection of subsets of $X$, that is $\tau\subset P(X)$.

We consider the collection $T$ of all topologies on X.

Pick a topology $\tau$ in $T$ one at a time, and let $(X,\tau)$ be the corresponding topological space.

If $(X,\tau)$ is a metrizable space, then let a corresponding metric space be $(X,d)$. Then, $|U_d|$ of the metric space $(X,d)$ is just equal to $|\tau|$. Note that there may be more than one metrics $d$ such that $(X,d)$ is a corresponding metric space to $(X,\tau)$.

If $(X,\tau)$ is not a metrizable space, then there is no metric $d$ such that $\tau$ is the induced topology by $d$, and hence for any metric $d$ on $X$ we will not have $\tau$ as a collection of open sets of a metric space $(X,d)$.

This gives an abstract algorithm to obtain the possible cardinality of a collection of all open sets of a metric space $(X,d)$ from an infinite set $X$.

Lastly, note that $|U_d|\leq |P(X)|$ for any metric $d$ on $X$.


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.