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Can anybody tell me how did submetrizability come into existence, and what is its use in topology ? Any examples to make me understand ?

Henno Brandsma
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2 Answers2

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There is a whole subbranch of general topology that came out of finding and proving metrisation theorems, and a lot of properties have been found that are so-called "Generalised Metric Spaces", that have nice properties and are "close to" metrisable in some sense. Monotonically normal spaces are such spaces (I recently talked about them here, in connection with the Michael line), developable spaces (introduced by Bing), spaces with a $G_\delta$ diagonal, $M$-spaces, $p$-spaces, etc. See Gruenhage's survey paper in the Handbook of Set theoretic Topology.

Many examples of such spaces (non-metrisable ones) do have the property, it was observed, that they have a natural weaker metrisable topology on the same set, e.g. the Sorgenfey line, the Michael line (which have the Euclidean topology as a weaker topology (a subtopology). So it is logical to start studying that feature as a new property in its own right, and call it submetrisable (British spelling). We can then prove interesting equivalent formulations of it (see Gruenhage's article) which make it look like other properties and prove some implications (like: submetrisable implies having a $G_\delta$ diagonal) so that it fits in a network of other properties. I cou;ldn't find who first introduced the property or its name (looks a bit like a property that Arhangel'skij would introduce to me) but it does fit into a wider family of properties.


Added based on comments: different guises under which we can meet this property:

TFAE:

  1. $(X, \tau)$ is submetrisable.

  2. There is a metric $d: X \times X \to \Bbb R$ (so a function obeying the metric space axioms) on $X$ that is continuous on $(X,\tau) \times (X,\tau)$.

  3. There is a continuous injection $f(X,\tau) \to (Y,d)$, where $(Y,d)$ is a metric space (of course having the metric topology $\tau_d)$.

$1 \to 2$: Let $\tau' \subseteq \tau$ be a topology on $X$ that is induced by some metric $d$ on $X$. Then $d$ is continuous as a map on $(X,\tau') \times (X, \tau')$, this is classical, and so the same holds for the stronger topology on $X^2$, $(X,\tau) \times (X,\tau)$.

$2 \to 3$: Let $Y=(X,d)$ and $f(x)=x$. $Y$ is by definition metric and $f$ an injection and $d$-open balls on $X$ are $\tau$-open as $B_d(x,r) = d_x^{-1}[(-\infty,r)]$, where $d_x: (X,\tau) \to \Bbb R, d_x(y)= d(x,y)$ is continuous as $d$ is. It follows that $f$ is continuous, as required.

$3 \to 1$: Given $f:(X, \tau) \to (Y,d)$ continuous and injective, define $\tau'$, a topology on $X$, by $\{f^{-1}[O]\mid O \in \tau_d\}$ and note that $\tau' \subseteq \tau$ and $(X,\tau') \simeq (f[X], d)$ so $X$ is submetrisable.


These are studied for being interesting, they have no "use" whatsoever. What's the use of an $L$-space? It's just expanding our knowledge of what's possible in topology.. I don't think it occurs as a condition in a theorem that is widely applied in analysis or elsewhere e.g. It's just one of the hundreds of properties that topologists have introduced for the sake of exploring. One interesting theorem (IMO): Every paracompact space with a $G_\delta$ diagonal is submetrisable. (If we replace paracompact by compact we get a metrisable space, then follows)

Henno Brandsma
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  • Thank you so much for all the efforts you have put to make me understand the concept. But as far as i have been through the definition of Submetrizability, there are two ways to define it: (1) A space having weaker metrizable topology. (2) A space $X$ is said to be submetrizable if there exists a continuous injection from $X$ to a metric space (say) $(Y, d)$. Could you please explain it to me that how these two definitions are equivalent? – Mir Aaliya Mar 06 '21 at 09:49
  • @HennoBrndsma The very thing i understood, but the rconverse i didn't get completely. One more thing, is it like there are two metrics on a space $X$ where one induces a topology on it and other does not. That is why is not metriozable with stronger topology(non induced by metric). – Mir Aaliya Mar 06 '21 at 11:08
  • @HennoBrndsma Let me confess what i understood is that: Let $X$ be a space and there is a topology $\tau_1$ on it with which it is metrizable. Then there is another topology $\tau_2 > \tau_1$ with which it is not metrizable. That means there is a metric $d$ on $X$ that induces the topology $\tau_1$ but not $\tau_2$. – Mir Aaliya Mar 06 '21 at 11:18
  • $\tau_2 \subseteq \tau_1$ and the metric induces $\tau_2$ but not $\tau_1$. It’s quite easy. Assuming $X$ comes with $\tau_1$ originally. – Henno Brandsma Mar 06 '21 at 11:23
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    @MirAaliya I added the arguments to the main post. – Henno Brandsma Mar 06 '21 at 11:57
  • @HennoBrndsma Got it. Thank you so much. – Mir Aaliya Mar 06 '21 at 17:17
  • @MirAaliya you’re welcome. Why do you need it? It does occur in some $C_p(X)$ theorems I believe. – Henno Brandsma Mar 06 '21 at 17:22
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I want to chip in to the nice answer by Henno Brandsma, with this small fact:

If $(X,τ)$ is submetrizable space, then any compact $K ⊆ X$ is metrizable.

Proof. Let $τ_d$ be a topology on $X$ inudced by a metric. Any compact $K\subset X$ is submetrizable as a subset of submetrizable space. We have $τ_d|K \subseteq τ|K$. Thus $K$ is compact in $τ_d|K$ as well (1). But $τ_d|K$, as a metric space, is Hausdorff and Hausdorff topology on a compact space is the largest possible (2). Thus $τ_d|K = τ|K$.

Kamil
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