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Among Metric Space,Uniform Space and Topological space which one is the strongest and who is the weakest $?$

As metric space is stronger than Topological space $\implies$ every metric can induce a topology i.e. every metric space is a topological space but not vice versa.There are many topological spaces that are not metrizable not first countable.First countablity and the existence of a metric are properties that a topological space may or may not possess.

My question is where uniform space fits in i.e. if it is at all comparable to those two.I thought it could be because it has the notion of open sets in it.Can we say metric $\implies$ uniform but not vice versa or something like this?

If we take a metric space $(X,d)$ and look at the product $X\times X$ can we provide a valid entourage here?If $\Phi$ is the entourage then it contains $\Delta=\{(x,x):x\in X\}$ i.e all open balls of the form $U\times U$ where $U$ is an open ball of $(X,d).$Now if that is not all we have in $\Phi$,say $U\times V\in \Phi$ s.t $U\neq V$ then how can $U\times V$ contain $\Delta$? But then if all we have is $U\times V$ in $\Phi$ then we cannot satisfy the second criterion of the definition. So it's not working out. That is we cannot say every metric space is uniform.

So at least uniform is weaker than metric.But how can we prove that every uniform space is metric space? What would be the $d$ here? Some hints please. Thank you.

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    $\begingroup$ You should see the example section on the wiki: en.m.wikipedia.org/wiki/Uniform_space. Every metric space is uniform, and certain Hausdorf uniform spaces are metrizable (not exclusively). Additionally, every uniform space can induce a topology. $\endgroup$ – David Jan 16 '17 at 9:20
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Every metric space has a uniformity induced by its metric, but not vice versa. Given $(X,d)$, define the set $$ E_\epsilon = \{(x,y) \in X^2: d(x,y) < \epsilon\} $$ Then $$ \Phi = \{ U \subseteq X| \exists \epsilon>0 : E_\epsilon \subseteq U\}$$ is a uniformity on $X$.

Hence, every metric space carries a canonical uniformity induced by its metric, but not vice versa: Some uniform spaces are "too big" to carry a metric, recall that there are arbitrary products of uniform spaces, the product $[0,1]^{[0,1]}$ cannot carry a metric, as its topology is not first countable, but it carries the product uniformity.

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Not every uniform space is metrizable. Precisely those that have countable base of entourages (and are separated) are. Uniformity is equivalent to having whole family of pseudometrics.

Also, every uniformity induces a topology: $\{E[x]: E$ is entourage$\}$ is a local base at $x$. But not every topological space is uniformizable. Precisely completely regular spaces are.

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  • $\begingroup$ So here it is : metric induces both topology and uniformity.Then uniformity indices topology but not vice versa. So topology<uniformity<metric. $\endgroup$ – user118494 Jan 16 '17 at 9:23
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    $\begingroup$ @user118494: Yes, that's it. But if you omit the symmetry axiom of uniformity, then every topological space is uniformizable. :-) $\endgroup$ – user87690 Jan 16 '17 at 9:25

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