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Consider the topological space $(\Bbb R,\mathfrak I)$ that arises from the metric space $(\Bbb R,d)$, with $d(x,y)=|x-y|$. I want to prove that $\partial(a,b)=\partial[a,b]=\{a,b\}$.

I have that

$x\in \partial A \iff d(x,A)=0\wedge d(x,X\setminus A)=0$

[This used to be a much longer and tortuous question, but since I can't delete it, I'll just leave what might interest other users, though it wasn't my main concern. When I find a suitable way to ask about my concern, I'll edit]

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    @ArturoMagidin I have defined a closed set as one whose complement is open. I guess that clears things up.2012-07-26

1 Answers 1

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If you want to use $\partial A = \overline{A}\cap \overline{X-A}$ you can proceed as follows:

  1. Step 1. If $x\in (a,b)$, then $x\notin\partial{(a,b)}$ and $x\notin\partial{[a,b]}$.

    Proof. Let $\epsilon = \min\{x-a,b-x\}$. Then $\{ y\mid d(x,y)\lt\frac{\epsilon}{2}\}\subseteq (a,b)$ and is open, so $x\notin\overline{X-(a,b)}$; since $\overline{X-[a,b]}\subseteq\overline{X-(a,b)}$, it follows that $x\notin\overline{X-[a,b]}$.

  2. Step 2. If $x\in (-\infty,a)$ then $x\notin\partial{(a,b)}$ and $x\notin\partial[a,b]$.

    Proof. Let $\epsilon =a-x$. Then $\{y\mid d(x,y)\lt \frac{\epsilon}{2}\}\subseteq (-\infty,a)\subseteq X-[a,b]$ and is open, so $x\notin\overline{[a,b]}$. Since $\overline{(a,b)}\subseteq\overline{[a,b]}$, it follows that $x\notin\overline{[a,b]}$.

  3. Step 3. If $x\in (b,\infty)$, then $x\notin\partial{(a,b)}$ and $x\notin\partial[a,b]$.

    Proof. Similar to that in step 2.

  4. Step 4. $a\in\partial(a,b)\cap\partial[a,b]$.

    Proof. Let $\epsilon\gt 0$. Let $\delta=\frac{1}{2}\min\{b-a,\epsilon\}$. Then $a+\delta\in (a,b)\cap \{x\mid d(x,a)\lt\epsilon\}$, and $a-\delta\in (X-[a,b])\cap \{x\mid d(x,a)\lt\epsilon\}$. Thus, every open ball containing $a$ intersects $(a,b)$, so $a\in\overline{(a,b)}\subseteq \overline{[a,b]}$; and every open ball containing $a$ intersects $X-[a,b]$ and so $a\in\overline{X-[a,b]}\subseteq\overline{X-(a,b)}$. Thus, $a\in \partial (a,b)\cap\partial[a,b]$.

  5. Step 5. $b\in\partial(a,b)\cap\partial[a,b]$.

    Proof. Similar to step 4.

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    @PeterTamaroff: "The elements" are the real numbers. The topology $\mathfrak{I}$ is defined *in terms* of the metric. Yes, you don't have to have the metric in mind to discuss the topology, *if* you already know enough about the topology. But for example, whenever you talk about the least upper bound "property", you are invoking the topology via the distance, so I'm not clear, again, what distinction you are trying to draw. That said, I'm going on hiatus, so I won't be able to reply in the future.2012-07-26