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


4 Answers 4


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!


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