0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ removed xxxxxxx $\endgroup$ – William Elliot Oct 24 '17 at 20:31
  • $\begingroup$ y in K subset U makes U a neighborhood. $\endgroup$ – William Elliot Oct 24 '17 at 20:47
  • $\begingroup$ 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 $\Bbb R.$ $\endgroup$ – DanielWainfleet Oct 25 '17 at 13:17
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.