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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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Do you want to only use the definition $\partial A=\overline{A}\cap\overline{X-A}$, or do you have some other eqeuivalences? –  Arturo Magidin Jul 26 '12 at 21:19
    
I'm not sure I like the argument much; how much is going into the assertion that the limit of a bounded sequence satisfies the same bounds, for example? –  Arturo Magidin Jul 26 '12 at 21:20
    
But how do you know that "limit point on the closed interval is contained in the closed interval"? That's essentially saying "closed intervals are closed", but it seems to me that this is precisely what you are trying to establish with the argument in the first place. –  Arturo Magidin Jul 26 '12 at 21:33
    
@ArturoMagidin I'm saying that since $a\leq x_n \leq b$. It follows that $a\leq \lim\;x_n \leq b$, which means $\lim\;x_n=x\in [a,b]$. I thought it was clear. –  Pedro Tamaroff Jul 26 '12 at 21:38
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Hmmm... If you have that exercise, why would you not want to use it? $d(a,(a,b))=d(a,[a,b]) = 0$, $d(a,X-(a,b))=d(a,X-[a,b]) = 0$, etc. Much less work and less danger of circularity. –  Arturo Magidin Jul 26 '12 at 21:48
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1 Answer

up vote 5 down vote accepted

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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In a first read, I don't follow your idea. Note that rather than asking for a completely different proof, I'm asking for feedback on my work. I know it might be a little long and farfetched, but it is what I could do. I'm a little concerned too with this: since $(\Bbb R,\mathfrak I)$ arises from $(\Bbb R,d)$, it suffices to prove the claim in the metric space, correct? See the add, for it might be relevant to this last question. –  Pedro Tamaroff Jul 26 '12 at 21:36
    
@PeterTamaroff: I don't understand what you mean by "proving the claim in the metric space". The metric space is the topological space: the topology is defined in terms of the metric. What is the difference you are trying to draw? –  Arturo Magidin Jul 26 '12 at 21:45
    
I thought $(\Bbb R,\mathfrak I)$ and $(\Bbb R,d)$ were not the same space. I thought $(\Bbb R,\mathfrak I)$ arose from $(\Bbb R,d)$ by "losing" the metric and retaining the open sets. It seems I was mistaken. That's why I talk about proving something is true for elements of $(\Bbb R,d)$ which would then mean it is also true for the equivalent elements of $(\Bbb R,\mathfrak I)$. I was thinking about equivalence rather than equality. –  Pedro Tamaroff Jul 26 '12 at 21:50
    
@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. –  Arturo Magidin Jul 26 '12 at 21:52
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