# Construct a metric that induces a given topology

In a topology textbook there was a exercise to determine the topology induced by $$x^2:\mathbb{R}\to\mathbb{R}$$ where the target has the euclidean topology.

I am the opinion that $$x^2$$ induced a kind of "mirrored" topology, meaning the open sets are all open sets of the euclidean topology that are symmetric at $$0$$.

However there was the bonus-question if this topology is induced by a metric. I figured out that this topology satisfy the second axiom of countability (take $$B=\{U_{1/n}(x)\mid x\in\mathbb{Q}, n\in\mathbb{N}$$}, so we can take the same set like in the euclidean topology) and this ensures that this topology comes from a metric.

If there are any errors I made please correct me. However, I now wondered if it would be possible not only to show that it comes from a metric, but also to explicitly give the metric that induces the topology? I tried some stuff out by unfortunately I was not able to do this, maybe anyone of you has an idea, or maybe it is not even possible?

I would appreciate any answers on this

More stronger(generalized) result:

Let $$(X, d)$$ be a metric space space and $$Y\subset X$$.

Let us consider a map $$f:X\to Y$$.

Let $$\tau_Y$$ be the topology induced by the function $$f$$ i.e $$\tau_Y=\{f^{-1} (U): U\in \tau_d\}$$

Then $$\tau_Y$$ is metrizable iff $$f$$ is injective.

Proof: Suppose $$f$$ is injective.

Then define the metric $$d_Y:Y\times Y\to X$$ by $$d_Y(y_1, y_2) =d(f(y_1, f(y_2))$$

Claim :(Left as an exercise)

1. $$d_Y$$ is indeed a metric on $$Y$$.

2. $$\tau(d_Y) =\tau_Y$$

Conversely : Suppose $$f$$ is not injective.

Then $$\exists y_1\neq y_2\in Y$$ such that $$f(y_1) =f(y_2)$$

Then $$\tau_Y$$ can't be Hausdorff.

For any set $$U\in \tau_d$$ containing $$f(y_1) =f(y_2)$$ contains both the points $$y_1$$ and $$y_2$$ .Hence this two distinct points $$y_1, y_2$$ can't be separated by two disjoint open sets. Hence $$\tau_Y$$ can't be Hausdorff and hence it can't be metrizable.

The topology is generated by a pseudometric $$d(x,y)=|x^2-y^2|$$ (which, of course, has nothing to do with the particular function or the eucldean distance on the range).

No it’s not induced by a metric. If would be so, the the space is Hausdorff, but if you take $$-1$$ and $$1$$ points, there are no disjoint open neighborhoods of them. In fact the image of $$-1$$ and $$1$$ under $$x^2$$ is $$1$$ and so neighborhoods of $$-1$$ and $$1$$ are inverse image of open neighborhoods of $$1$$, that contains always both $$-1$$ and $$1$$.

By your observation, neighborhoods of $$1$$ in this topology are symmetric about $$0$$ so $$(-1,-1,-1,\cdots) \to 1$$ in this topology. If the topology is given by a metric $$d$$ then we get $$d(-1,1)=0$$ a contradiction. Actually, this topology is not even Hausdorff!