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I am assuming the $\Bbb{R}$ that $f$ maps to is equipped with the natural topology.

The induced topology is $T_f=\{ f^{-1}(I) | I \subset \Bbb{R} \text{ open}\}$. Let $I \subset \Bbb{R}$ be open in the natural topology, then $I$ can be expressed as a union of open intervals $(a,b) \subset \Bbb{R}$. Those open intervals have preimages consisting of open intervals and their "mirrored" counterparts, $(c,d)$ and $(-d, -c)$.

So $T_f$ consists of unions of such interval pairs. But those are precisely the pairs $I, -I$ for $I \subset \Bbb{R}$ open in the natural topology.

Can $T_f$ be induced by a metric? Intuitively, I think the answer is no, because the ball of radius $0$ around $x \in \Bbb{R}$ would contain $x$ aswell as $-x$. Is this correct?

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The topology is not Hausdorff hence not metrizable. Note that any open set containing $1$ also contains $-1$ which proves that the topology is not Hausdorff.

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It is not metrizable because it isn't T0. In fact, points $x$ and $-x$ are never topologically distiguishable in $T_f$.

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