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This question already has an answer here:

I wanted to know if there is a distance function $d$ on $\Bbb R$ so that a nonempty subset $U$ of $\Bbb R$ is open with respect to $d$ if and only if its complement $\Bbb R$ \ U is finite ?

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marked as duplicate by user21820, user99914, Namaste, Asaf Karagila general-topology Jul 29 '18 at 15:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Perhaps overkill, but here is another proof of the impossibility of such a metric:

You can prove that in any metric space $X$ and for all $x \in X$, $$\bigcap_{n=1}^\infty B_\frac{1}{n}(x)= \{x\}$$ (where $B_\epsilon(x)$ is the open ball of radius $\epsilon$ centered at $x$).

Suppose that you had such a metric on $\Bbb R$ and fix an $x \in \Bbb R$. By assumption, the complement of every open ball centered at $x$ is finite. Thus

$$\Bbb R = \{x\} \cup \bigcup_{n=1}^\infty B^{\,\prime}_\frac{1}{n}(x)$$ is a countable union of finite sets, hence countable. But this contradicts the uncountability of $\Bbb R$.

More generally, the cofinite topology on an uncountable space isn't first-countable hence it isn't metrizable since all metric spaces are first-countable.

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No, this is not possible. The resulting topology is not Hausdorff, and any metric topology must be.

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Let $d$ be such a metric and $a,b$ any distinct points. Then the open balls around $a$ and $b$ of radius $\frac 12 d(a,b)$ are disjoint proper non-empty open sets, hence at most one of them can be co-finite.

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