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Given $(X,d)$, I need to show the above statement. I found a question here that I initially thought would answer my query

Showing that metric induces single unique topology on a finite set

However, I understand this proves that a metric on a finite space can induce the discrete topology

I don't see why this is the only(exactly one) topology on $X$

So I am stuck with reasoning why this is unique. What assures me that there is absolutely no other topology thinkable with any given metric? Just because I've found that it would induce a particular topology, nothing should be stopping me so far from finding another.

Well, an anticipated comment would be that "well you can try if you think so" but that's not what I'm getting at; I might not be able to "think" of one, I am looking to reason "this is impossible."

Can someone concisely explain uniqueness

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  • $\begingroup$ The question is not saying that it's the only possible topology one can put on $X$ (one could always use the indiscrete topology), it's that given the topology induced by a metric on a finite set, that topology must be the discrete topology $\endgroup$ – leibnewtz May 6 '16 at 17:10
  • $\begingroup$ Yes, well my question is why is that the ONLY topology induceable by a metric on a finite space. $\endgroup$ – John Trail May 6 '16 at 17:11
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    $\begingroup$ No the proof given shows that any singleton must be open in the induced topology. This can only happen if the topology is discrete $\endgroup$ – leibnewtz May 6 '16 at 17:12
  • $\begingroup$ The comments in the question address the same problem by the way $\endgroup$ – leibnewtz May 6 '16 at 17:13
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    $\begingroup$ The proof shows that in the topology induced by an arbitrary metric, any singleton is open. Now suppose some topology induced by some metric is not discrete. Since every singleton is open however (for any metric), any union of those singletons is open, implying that any subset is open, contradicting our assumption that the topology was not discrete. It follows that any topology induced by a metric must be discrete $\endgroup$ – leibnewtz May 6 '16 at 17:29
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The topology induced by a metric space $(X,d)$ is defined as the smallest topology $\mathcal{T}$ on $X$ such that all balls $B(x,r) = \{y \in X: d(x,y) < r \}$ are in $\mathcal{T}$ for all $x \in X$ and all $r>0$ in the reals.

Formally, this would be the intersection of all topologies on $X$ that have the property that they contain all balls $B(x,r)$. There is at least such topology on any set $X$, the discrete topology $\mathcal{T}_{\text{disc}} = \mathcal{P}(X)$ on $X$, where all subsets of $X$ are in the topology (so in particular all balls). So this topology is well-defined as this intersection.

Now, the proof here shows that if $X$ is a finite set, then for every $x \in X$ there is some radius $r_x > 0$ such that $B(x,r_x) = \{x\}$. This shows in turn that a topology $\mathcal{T}$ that contains all balls, contains all sets $\{x\}$, $x \in X$, and so in fact all subsets $A$ of $X$, as $A = \cup\{\{x\} : x \in A\}$ by definition, and as all topologies are closed under unions (which are here even finite). Alternatively, using the well-known fact that singletons are closed in metric spaces, all sets are closed as finite unions of singletons, and so all sets are open as well (as their complement is closed).

Hence, there is only one topology on $X$ that contains all balls: namely all subsets of $X$, or the discrete topology. So the induced topology is the intersection of a single topology $\mathcal{T}_{\text{disc}}$.

The confusion seems to arise from the idea that there could be more than one topology induced by a metric, but per definition this cannot be, as it is uniquely defined as the above-mentioned intersection. It is not the unique topology on $X$, but it is the unique topology that can be induced by any metric (such topologies are quite special among all topologies on a set).

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