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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?

Polymorph
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1 Answers1

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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 $Y$.

If you're allowed to, you should also look into verifying the subspace topology is a topology using the usual axioms involving open sets; I think that's much more natural.

yandz
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