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I wanted to prove $\partial \bar N\subset \partial N$ in a different way, by showing if $x \in \partial \overline{N}$ then $x \in \partial N$. This I wanted to show by showing that if $U$ is an nbhood of $x$ then it contains

(i) a point of $N$ and

(ii) a point of $N^c$.

Since $U$ is an nbhood of $x$ it contains a point of $\overline{N}$ and a point of $\overline{N}^c$. We have $N \subset \overline{N}$ hence $\overline{N}^c \subset N^c$ hence (ii) is clear. What I'm stuck with is (i). I tried the following: $U$ contains a point $y$ of $\overline{N} = \partial N \cup \mathrm{int}N$. If $y \in \mathrm{int}N$ then we're done. If $y$ in $\partial N$ then either $y \in N$ or $y\in N^c$. If $y \in N$ we're done.

But if $y \in N^c$ then... nothing. So it looks as if $U$ could be a subset of $N^c$. Is it really not possible to prove it like this or do I just not see how? Thanks for help.

Community
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Rudy the Reindeer
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  • Since $\bar{N}$ is closed it contains its boundary. Hence an element of $\partial\bar{N}$ is also an element of $\bar{N}$. Shouldn't this be sufficient for (i)? – M.B. Oct 18 '12 at 13:05

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Clarification: In this post whenever I say the word neighborhood I mean an open neighborhood. (Based on the comments, it seems that this caused some misunderstandings. And I have to admit that I did not think about this when posting.)

If $x\in\overline N$ if and only if every neighborhood $U$ of $x$ contains a point from $N$.

$\boxed{\Rightarrow}$ Suppose that there is a neighborhood $U$ such that $U\cap N=\emptyset$. Then $N\subseteq X\setminus U$ and $X\setminus U$ is closed, hence $\overline N\subseteq X\setminus U$. I.e., $\overline N\cap U=\emptyset$, contradicting the assumption that $x$ is in this intersection.

$\boxed{\Leftarrow}$ If $x\notin\overline N$ then $U:=X\setminus\overline N$ is a neighborhood of $x$ such that $U\cap N=\emptyset$.

From this you can easily see that if every neighborhood of $x$ contains a point from $\overline N$, then every neighborhood of $x$ contains a point from $N$.

(Let $U$ be a neighborhood of $x$. If $U\cap\overline N\ne\emptyset$, then we have a point $y\in U\cap \overline N$. Since $U$ is a neighborhood of $y$ and $y\in\overline N$, we get that $U\cap N\ne\emptyset.$)

Martin Sleziak
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  • In your last sentence, $U$ is a nbhood of $x$. Not necessarily of $y$. Or do we by convention assume nbhoods to be open? – Rudy the Reindeer Oct 18 '12 at 13:12
  • You are correct, I should have written open neighborhood everywhere. – Martin Sleziak Oct 18 '12 at 13:14
  • I don't know. I'm having a tired day. I'll re-read your answer in a few hours. – Rudy the Reindeer Oct 18 '12 at 13:25
  • Dear Martin, I think my previous comment is irrelevant. However, regarding the last line of your answer: why is $U$ a neighbourhood of $y$? We assume it to be a neighbourhood of $x$. We need to argue why there is an open set $O$ such that $y \in O \subset U \cap \overline{N}$. – Rudy the Reindeer Oct 30 '12 at 07:58
  • @MattN Would you agree that it is ok, if we know that $U$ is an open neighborhood? In the whole post I should replace neighborhood by open neighborhood - it would make it more clear. – Martin Sleziak Oct 30 '12 at 07:59
  • No. If $U$ is open then why is $U \cap \overline{N}$ open? – Rudy the Reindeer Oct 30 '12 at 08:00
  • But I don't need to show that $U\cap\overline N$ is open. (It is not even true in general.) I only need to show that $U\cap N$ is non-empty. – Martin Sleziak Oct 30 '12 at 08:04
  • We want to show that $U \cap N \neq \varnothing$. We want to do this by using your lemma above. We have $y \in U \cap \overline{N}$. Hence $y \in \overline{N}$. Now if we can show that $U \cap N$ is a nbhood of $y$ then we can apply your lemma and are done. But why is $U \cap N$ a nbhood of $y$?(I'm sorry for the typo in my previous comment.) – Rudy the Reindeer Oct 30 '12 at 08:47
  • It suffices to use te fact that $U$ is an open neighborhood of $y$. Then (using the lemma) $y\in\overline N$ implies $U\cap N\ne\emptyset$. – Martin Sleziak Oct 30 '12 at 08:51
  • Thank you, I'm getting there. And why is $U$ a neighbourhood of $y$? (oh dear...) – Rudy the Reindeer Oct 30 '12 at 08:55
  • Well, it is an open set and it contains $y$. (This is why I have to work with open neighborhoods, so that this argument works.) – Martin Sleziak Oct 30 '12 at 08:58
  • Ouch x_x Thank you for your patience. – Rudy the Reindeer Oct 30 '12 at 08:59