Is a symmetric premetric space a topological space

general-topology

Let us recall… the metric

A metric is a function $d$ over the Cartesian product of elements of a set and with a real number as output.
The set and the metric are called a "metric space".
For $d$ it must apply:

  1. $d(x,y)\geq 0$ non-negativity or separation axiom
  2. $d(x,y)=0\Leftrightarrow x=y$ identity of indiscernibles
  3. $d(x,y)=d(y,x)$ symmetry
  4. $d(x,y)\leq d(x,z)+d(z,y)$ subadditivity or triangle inequality

What I am given… a symmetric premetric

I am given a symmetric premetric $d$, so a function $d: x,y \rightarrow \mathbb R$ for which holds

  • $d(x,y)\ge0$
  • $x=y \Rightarrow d(x,y)=0$
  • $d(x,y)=d(y,x)$,

where $x, y \in X$, where $X$ is a set.
Compared to a metric I am missing the requirements

  • $d(x,y)=0 \Rightarrow x=y$
  • $d(x,y)\leq d(x,z)+d(z,y)$ subadditivity or triangle inequality

Questions:

  1. Is the space $(X, d)$ a topological space?
  2. Is it called "premetric space" or "symmetric premetric space"?

Trying to answer the question myself…

The only definition of topological spaces that I get a little bit, is the one of Felix Hausdorff:

  1. If N is a neighbourhood of x (i.e., N ∈ N(x)), then x ∈ N. In other words, each point belongs to every one of its neighbourhoods.
  2. If N is a subset of X and includes a neighbourhood of x, then N is a neighbourhood of x. I.e., every superset of a neighbourhood of a point x in X is again a neighbourhood of x.
  3. The intersection of two neighbourhoods of x is a neighbourhood of x.
  4. Any neighbourhood N of x includes a neighbourhood M of x such that N is a neighbourhood of each point of M.

I think that the neighborhoods can be ordered according to $d$ and that for two neighborhoods $N_i$ and $N_j$ all elements of one of them must be completely contained within the other (or vice versa or both).

  1. is the case: The "smallest neighborhood" for $x$ would be the set of all $y$ for which $d(x,y)=0$, which contains $x$.
  2. is the case due to the ordering.
  3. is the case due to the ordering, so the intersection is the "smaller" neighborhood of the two would be the intersection.
  4. is the hardest: I am not sure what $M$ would be, except for $M=\{x\}$. Honestly, I would not be able to answer this differently for an Euclidean metric space either.

Answer (for myself and whoever is interested)

Reading Henno Brandsma's answer and discussing with him helped me to get the following answer, that I would like write down, so that I have it in my own words. Maybe it helps others, too.

First of all, $(X, d)$ is not a topological space, because a topological space is a set with a topology and $d$ is a premetric, not a topology.

As we know from https://en.wikipedia.org/wiki/Topological_space:

In topology and related branches of mathematics, a topological space may be defined as a set of points, along with a set of neighbourhoods for each point, satisfying a set of axioms relating points and neighbourhoods.

So, since neighborhoods are also sets, the topology is a "set of sets, satisfying specific axioms".

Now, my original thought was, that the $d$ directly defines these neighborhoods with, meaning that the set of balls $B_d$, based on $d$, namely

$$
\tau' = \{ B_d(x, r) | x\in X, r\in\mathbb R, r>0 \}
$$

is also not a topology, because it does not fulfill the last condition by Hausdorff. If we are considering the other definition for a topology $\tau$, namely that

$\tau$ is a collection of subsets of $X$, satisfying that

  1. The empty set and X itself belong to $\tau$.
  2. Any arbitrary (finite or infinite) union of members of $\tau$ still belongs to $\tau$.
  3. The intersection of any finite number of members of $\tau$ still belongs to $\tau$.

we see that $\tau'$ does not fulfill the conditions 2 and 3.

Instead, we can say that $\tau'$ is a base for a topology $\tau$. We get $\tau$ taking $\tau'$, but all it's elements into the set $\tau$ and also all unions and intersections of the elements of $\tau'$. Now, for each element $O$ of $\tau$, we can show what Henno Brandsma showed in his answer and thus show that $\tau$ is a topology and $(X, \tau)$ a topological space.

This topological space $(X, \tau)$ is in fact induced by a premetric space $(X, d)$, but that does not mean, that $(X, d)$ is a topological space, or that $(X, \tau')$ is a topological space. This also answers my question 2. In fact, I found the term premetric space in publications.

Best Answer

The usual way to define a topology $\mathcal{T}_d$ from a metric-like function $d: X \times X \to \Bbb R$ is to define $B_d(x,r)=\{y \in X: d(x,y) < r\}$ for $x \in X$ and $r>0$.

$O \subseteq X$ is then called open iff $$\forall x \in O: \exists r>0: B_d(x,r) \subseteq O\tag{1}$$

Checking the usual axioms for open sets:

$X$ is open is trivial, for any $x \in X$ we can take $r=1$ (or whatever) to fulfill $(1)$. $\emptyset$ is open because there are no $x$ in it to check $(1)$ on (void truth).

If $O_i, i \in I$ is a family of open sets, then $O=\bigcup_{i \in I}O_i$ is open: if $x \in O$, then for some $i_0 \in I$ we have $x \in O_{i_0}$. As that set is open by $(1)$ we have $r>0$ such that $B_d(x,r) \subseteq O_{i_0}$. Because $O_{i_0} \subseteq O$ (as always for unions) that same $r$ works to fulfil $(1)$ for $O$ and $x$. So $O$ is open.

If $O_1$ and $O_2$ is open, let $x \in O_1 \cap O_2$ be arbitrary. $x \in O_1$ gives us $r_1>0$ such that $B_d(x, r_1) \subseteq O_1$ and $x \in O_2$ gives us $r_2>0$ such that $B_d(x, r_2) \subseteq O_2$. Set $r=\min(r_1,r_2)$ and regardless of any axioms on $d$ we know that $$d(x,y) < r_1 \land d(x,y) < r_2 \iff d(x,y) < r$$ It follows that $$B_d(x,r) \subseteq O_1 \cap O_2$$ and $(1)$ is fulfilled for $x$ and $O_1 \cap O_2$. So $O_1 \cap O_2$ is open.

So this also defines a natural topology for a symmetric premetric $d$. If $d$ is a full metric, we get the standard metric topology.

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