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Is there a distance $d$ such that the metric topology on $\mathbb{R}$ is the set of symmetric intervals, $T(d) = \{(-a,a),[-a,a] | a\in \mathbb{R}\}\cup \mathbb{R}\cup \emptyset$?

The definition that I have for metric topology is $T(d) = \{U \subset \mathbb{R} : \forall p \in U:\exists \epsilon > 0 :B_{\epsilon}(p)\subset U\}$.

I don't think such a distance exists. In such a topology $U$ non-empty, open $\implies 0 \in U.$

I'm looking for a contradiction and I have a gut feeling that it has something to do with this set.

$[0,0] = \{0\} \in T(D)$, so $\exists \epsilon > 0$ st $B_{\epsilon}(0)\subset \{0\}$, but $0\in B_{\epsilon}(0)$ so $B_{\epsilon}(0) = \{0\}$.

And now I am stuck for ideas.

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    $\begingroup$ That wouldn't be Hausdorff. You have that $1$ and $-1$ can't be separated by open sets. $\endgroup$
    – J126
    Jan 2, 2018 at 14:03
  • $\begingroup$ Ah yes, I see. Any two non-empty open sets can't have an empty intersect and every metric space is Hausdorff. $\endgroup$ Jan 2, 2018 at 14:11

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Indeed, notice that for every $\varepsilon > 0$ the open ball $B_\varepsilon(1)$ contains $0$, because every non-empty open set contains $0$.

Therefore, $d(0, 1) < \varepsilon$, for all $\varepsilon > 0$. Hence $d(0, 1) = 0$, which is a contradiction.

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The reason you gave, namely that for all non-empty open sets $O$, we have $0 \in O$ is a valid way to see the space is not metrisable:

Suppose $d$ existed. Then $r= d(1,0) >0$ and the ball $B(1, r)$ is non-empty (contains $1$) and does not contain $0$ (we don't have $d(1,0) < r$ obviously) contrary to the above observation.

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