# Can the topology induced by $f: \Bbb{R} \rightarrow\Bbb{R}, f(x)=x^2$ be induced by a metric?

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?

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