Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $T$ a topological space and $S\subseteq T$ a subspace equipped with the subspace topology inherited from $T$. Take a subset $H\subseteq S$. I'd like to prove that $\partial_S H=S\cap \partial_T H$ (where $\partial_X A$ denotes the boundary of $A$ in $X$) but I suspect that this is true only if $S$ is open in $T$. First I've shown $\partial_S H\subseteq S\cap \partial_T H$ as follows: $x\in \partial_S H$ if and only if for every (open) neighbourhood $U\subseteq S$ of $x$ both $U\cap H\neq\emptyset$ and $U\cap H^c\neq\emptyset$ hold. Now, if $V$ is open in $T$ then $U=V\cap S$ is open in $S$ and $U\cap H, U\cap H^c\neq\emptyset$, so in particular $V\cap H, V\cap H^c\neq\emptyset$.

For the other inclusion, take $x\in S\cap\partial_T H$. Then, if $U=V\cap S$ is open in $S$ (where $V$ is open in $T$), we'd like to say that $U\cap H\neq\emptyset$ (similarly for $H^c$), but we should use that U is open also in $T$, which is true if $S$ is open in $T$.

Now, if this proof is correct, I am through for $S$ open in $T$. What can we say otherwise? Are the corresponding statements for the interior and the closure of $H$ still true? (I guess so, with an almost identical proof!)

Thanks, bye!

share|improve this question
add comment

2 Answers 2

Consider the case where $S$ has nonempty boundary in $T$ that intersects $S$, and let $H=S$. Then trivially $\partial_S H = \partial_S S = \emptyset$, but $\partial_T S\cap S \neq\emptyset$.

For example, take $S=[0,1]$ in $T=\mathbb{R}$ and $H=[0,1]$. Then $\partial_T H = \partial_T([0,1]) = \{0,1\}$.

However, $\partial_S H = \emptyset$, because no $S$-neighborhood of $H$ intersects the complement of $H$ in $S$.

If you want an example with $H\neq S$, then just take $H=(0,1]$ and $S$, $T$ as before. Then $\partial_T(H)\cap S = \{0,1\}$, but $\partial_S(H) = \{0\}$.


On the other hand, it is always true that $\partial_S H\subseteq \partial_T H\cap S$:

Let $x\in \partial_S H$, and let $U$ be an open set of $T$ that contains $x$. Then $U\cap S$ is open in $S$ and contains $x$, hence $$\varnothing \neq H\cap (U\cap S) = (H\cap S)\cap U = H\cap U.$$ So $H\cap U\neq\varnothing$. And $$\varnothing \neq (S-H)\cap (U\cap S) \subseteq (T-H)\cap U,$$ hence $U\cap (T-H)\neq \varnothing$. Hence $x\in \partial_T H$, as claimed.


You should be really careful with the complement: note that you have two different notions of "$H^c$" at play in the situation above: there's the complement of $H$ in $S$, and the complement of $H$ in $T$. It's probably better to use $S-H$ and $T-H$ in your arguments, to avoid possible confusion (though, of course, $S-H\subseteq T-H$).

share|improve this answer
add comment

By definition the set $S$ viewed in its subspace has no boundary points. Viewed in the setting of its parent space $T$, it can have boundaries : see this diagram.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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