Number of subsets/open subsets/closed subsets of a metric space.

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 $d$?

• 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.
– Ian
Aug 16 '14 at 23:36
• @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 $d$?
– User
Aug 17 '14 at 1:16
• 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 $z \in Z$.
– 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 $X$.