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I am trying to prove the following statement regarding nowhere dense sets:

"In a metric space X, the frontier of an open set is the set of accumulation points of a discrete set."

As far as my attempts go, I have gone back and looked at an earlier proof of Baire's Category Theorem and worked out the details (as to my understanding, it should be similar). However, I am still not seeing how to prove the above using said theorem. Does anyone have any hints or suggestions as to how to proceed?

Note: I have already proven (as part of the above) that:

a)In a metric space X without isolated points, the closure of a discrete set in X is nowhere dense in X.

b) In any space X, the frontier of an open set is closed and nowhere dense.

c) Every closed nowhere dense set is the frontier of an open set.

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    $\begingroup$ Where does nowhere dense occur in this? And aren't ther simple counterexamples to the claim as is? $\endgroup$ – Hagen von Eitzen Feb 8 '16 at 7:33
  • $\begingroup$ This is part of a 4-part exercise in S. Willard's book "General Topology" found on pg. 37, for which one of the parts asks to prove that "In any space X, the frontier of an open set is closed and nowhere dense." The converse is also proven. Not sure as to counterexamples to this claim, it seems to be true to me. I will edit the post with this info. $\endgroup$ – LordVader007 Feb 8 '16 at 7:39
  • $\begingroup$ Regarding counterexamples: With $X=\Bbb C$, how is $\Bbb R$ (being the boundary of the upper half plane) the closure of a discrete set? $\endgroup$ – Hagen von Eitzen Feb 8 '16 at 7:54
  • $\begingroup$ @Hagen: Let $$D=\bigcup_{n\in\Bbb N}\left\{(k+i)2^{-n}:k\in\Bbb Z\right\}\;;$$ $D$ is discrete, and its set of accumulation points is the real axis. $\endgroup$ – Brian M. Scott Feb 8 '16 at 8:35
  • $\begingroup$ @BrianM.Scott Ah, I somehow read closure instead of accumulation points $\endgroup$ – Hagen von Eitzen Feb 8 '16 at 10:28
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As Willard says, this is a little tricky. The key idea can be seen in my comment to Hagen von Eitzen above.

HINT: First prove the following lemma.

Lemma. Let $\langle X,d\rangle$ be a metric space, let $U$ be a non-empty open set in $X$, and let $\epsilon>0$. Then there is a discrete set $D_\epsilon(U)\subseteq U$ such that $\bigcup_{x\in D_\epsilon(U)}B(x,\epsilon)\supseteq U$.

To do so, apply Zorn’s lemma to the family of subsets $A$ of $U$ such that $d(x,y)\ge\epsilon$ whenever $x$ and $y$ are distinct points of $A$.

Now for $n\in\Bbb N$ let

$$U_n=\left\{x\in U:2^{-n-1}<d(x,X\setminus U)<2^{-n}\right\}\;,$$

and consider the set

$$D=\bigcup_{n\in\Bbb N}D_{2^{-n}}(U_n)\;.$$

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  • $\begingroup$ Thanks for the hint, Brian. However, I am still a bit confused about your notation regarding the discrete set above. In particular, what does $D_\epsilon(U)\subseteq U$ stand for? $\endgroup$ – LordVader007 Feb 9 '16 at 4:05
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    $\begingroup$ @LordVader007: It’s defined in the statement of the lemma: it is some discrete subset of $U$ with the property that every point of $U$ is contained in the $\epsilon$-ball centred at some point of $D_\epsilon(U)$. The lemma is simply the assertion that for each $\epsilon$ and $U$ such a set exists. $\endgroup$ – Brian M. Scott Feb 9 '16 at 4:22

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