# Verification of the subspace topology defined in terms of neighborhoods

I'm currently trying to verify that the subspace topology is indeed a topology as defined in this post (Verifying the subspace topology with a topology defined in terms of neighborhoods). My problem is with the verification of Axiom c).

Let X denote the space and Y the subspace.

If I assume that $U \subset Y$ is a neighborhood of y with respect to Y then by definition of the subspace topology there is a neighborhood K of y with respect to X such that $U=Y\cap K$.

Suppose that there is $\ L \subset Y$ such that $U \subset L.$

It is obvious that $L=Y\cap L$. Now I would like to prove that $U=Y\cap K$ is a neighborhood of y with respect to X and then use the fact that X is a topological space (for which the axioms a)-d) hold) to conclude that L is a neighborhood of y with respect to X. This would allow me to deduce that L is a neighborhood of y with respect to Y and finish the proof.

But I do not see how to prove that U is a neighborhood of y with respect to X at all. Is this even possible? Or is there some other approach?

• removed xxxxxxx – William Elliot Oct 24 '17 at 20:31
• y in K subset U makes U a neighborhood. – William Elliot Oct 24 '17 at 20:47
• If $U$ is a nbhd of y with respect to $Y$ then $U$ need not be a nbhd of $y$ with respect to $X.$ For example if $Y=\Bbb Q$ and $X=\Bbb R$ with the usual topology on $\Bbb R.$ – DanielWainfleet Oct 25 '17 at 13:17

Suppose $U \subset Y$ is a neighborhood of $y$ wrt the subspace topology. Then $U$ is also a neighborhood of $y$ in $X$ (there's no reason for the axioms to be false in $X$ if they are true in $Y$). Take $L$, a subset of $Y$ (and thus also a subset of $X$) containing $U$. Clearly, $X \supset L \supset U$ in $X \implies L$ is a neighborhood of $y$ in $X$. Since the neighborhoods in a subspace are just intersections of neighborhoods in $X$ with $Y$, we have $L = L \cap Y$ is also a neighborhood of $y$ in $Y$.