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The problem: Come up with an example of sets $A$, and $B$ in $\mathbb{R}^2$ such that $A \subset B$, $A \neq B$, and the boundary points of $A$, $bd(A) = B$.

Here is what I think. I was going to set $A = \{ \bar{u} \in \mathbb{R}^2 | \|u \| =1 \} $ and $B = \{ \bar{u} \in \mathbb{R}^2 | \| u \| \leq 1 \}$ but I cannot think of good examples.

Can someone give me some hints please? Try not to solve the problem!

Thank you very much!!

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    Perhaps you mean $B\subset A$.? – 2017-02-21
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    @MyGlasses I meant A is a subset of B – 2017-02-21
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    @Vikrant Desai, I actually made a correction. The third condition should be that $bd(A) = B$. – 2017-02-21
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    Then your example is wrong. Becasue $bd(A)=A \subset B$. – 2017-02-21
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    Let $A = \{(1/n,0):n \in \mathbb N\},B = A \cup \{(0,0)\}.$ – 2017-02-21
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    $A=\{(x,0):0\lt x\lt1\},\ B=bd(A)=\{(x,0):0\le x\le1\}.$ – 2017-02-21

3 Answers 3

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Take $B = \{ \bar{u} \in \mathbb{R}^2 | \|u \| =1 \}$ and remove any countable number of points you wish from $B$ to get $A$.

By countable, I mean finite or countably infinite.

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$A = \{ u=(x,y) \in \mathbb{R}^2 | \| u \| \leq 1,x,y\in\mathbb{Q} \}$

$B = \{ u \in \mathbb{R}^2 | \| u \| \leq 1 \}$

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Your example does not work, since $(0,0)$ is not a boundary point of B, just take the ball of radius $\frac{1}{2}$ around it, which clearly does not contain any elements from $A$. You need something dense in $B$ to be your set, and there is nothing quite like $\mathbb{Q}$ as far as dense sets go. Can you finish it from here?

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    Do you mind if I ask you another question? It has to do with set theory like this. – 2017-02-22
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    Sure. What is your question. – 2017-02-22
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    Here is the question: Let $A$ and $B$ be sets in $\mathbb{R}$ so that $A \subset B$, $A \neq B$, $bd(A) \subset B$ and B is closed in $\mathbb{R}$. Come up with an example of two sets which show that "$B$ is closed " is a neseccary condition. So show examples which $A \subset B$, $A \neq B$, $bd(A) \subset B$, and $B$ is not closed. What I have as my examples is when $A = \mathbb{Q}$ and $B = (-\infty, 0) \cup (0, \infty).$ However, this is not a good example and I am stumped on coming up with one. – 2017-02-22
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    This would actually be better as a separate posted question. It is pretty hard to read it when the Tex is not compiled. – 2017-02-22