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My lecturer says as follows; $C_0 = \{(a,b) : -\infty \le a\le b< \infty\}$

$$C_\mathrm{open} = \{ A \in \mathbb{R} : A\text{ open} \}$$

He goes on to show that $\sigma(C_0) = \sigma(C_{\mathrm{open}})$;

Clearly, $\sigma(C_0)$ is in $\sigma(C_{\mathrm{open}})$

So now I need to show the other way round, $\sigma(C_{\mathrm{open}})$ is in $\sigma (C_0)$

I do this by showing that $C_{\mathrm{open}}$ is in $\sigma (C_0)$

He says; take $A$ as subset of $\mathbb{R}$ which is open, then $$A= \bigcup_{x \in X} (x-\varepsilon_x, x + \varepsilon_x)$$ Then he says $A \cap\mathbb{Q}$ is a subset of $$A= \bigcup_{x \in X} (x-\varepsilon_x, X + \varepsilon_x)$$ and $A\cap\mathbb{Q} = \{y_1,\ldots\}$ such that there exists $x_n$ s.t $y_n$ is in $(x_n-\varepsilon_{x_n}, x_n+\varepsilon_{x_n})$ for all $n$ of course $$\bigcup_{n=1}^{\infty} (x_n-\varepsilon_{x_n}, x_n+\varepsilon_{x_n})$$ is a subset of $A$

Then he says let $$\varepsilon_{n-} = \sup\{\varepsilon>0\mid (x_n-\varepsilon, x_n] \subseteq A\}$$ and let $$\varepsilon_{n+} = \sup\{\varepsilon>0\mid [x_n, x_n+\varepsilon)\subseteq A\}$$

Now we need to show that $A$ is a subset of $$\bigcup_{n=1}^{\infty}(x_n-\varepsilon_{n-},x_n+\varepsilon_{n+})$$

Take $x$ in $A$ then as before $x$ is in $(x-\varepsilon_x, x + \varepsilon_x)$. $\mathbb{Q}$ is dense, so there exists $n$ s.t $y_n$ is in $(x-\varepsilon_x, x + \varepsilon_x)$ which is a subset of $(x_n-\varepsilon_{n-},x_n+\varepsilon_{n+})$ by definition of $\varepsilon_{n+}$ and $\varepsilon_{n-}$.

This then implies that $A$ is a subset of $$\bigcup_{n=1}^{\infty}(x_n-\varepsilon_{n-} , x_n+\varepsilon_{n+})$$

So our arbitrary $A$ is in $\sigma(C_0)$

Can anybody explain this to me, I don't understand where the steps come from and lead to. All I know is you need to use the rationals because they're countable, unlike the reals.

Thanks

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Dear Rosie, Welcome to math stackexchange. Kindly refer here (meta.math.stackexchange.com/questions/107) on how to typeset (i.e. how to write equations etc) on this website. This is done so that the equations and other math appear in a nice way. –  user17762 May 20 '12 at 22:12
    
@Rosie: I attempted to format your question; please review to make sure I did not change the meaning. Also, you don't need to include your name: it is posted at the bottom right automatically. –  Arturo Magidin May 20 '12 at 22:29
    
Sorry, I'm new to this and didnt realise you could do the equations like that. Thank you @Marvis –  Rosie May 20 '12 at 22:34
    
@Rosie Thats fine. I just want to make you aware of the way to write math on this site. –  user17762 May 20 '12 at 23:30

2 Answers 2

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What your instructor does is the following: He tries to write $A$ as a countable union of open intervals. He first writes $A$ as a uncountable union of intervals $(x-\varepsilon_x, x + \varepsilon_x)$. He then picks a rational number $y_n$ from each interval and picks then a point $x_n$ such that $y_n\in(x_n-\varepsilon_{x_n}, x_n+\varepsilon_{x_n})$. This step is actually superflous, one can just take $x_n=y_n$. He then puts a maximally large open intervall around $x_n$. This is where the $\varepsilon_{n-}$ and $\varepsilon_{n+}$ come in. They are constructed so that $(x_n-\varepsilon_{n-},x_n+\varepsilon_{n+})$ is the largest possible open interval around $x_n$ that lies still in $A$. Then he verifies that the union of these sets coincides with $A$. The motivation that this is a useful approach comes from the fact that every open subset of $\mathbb{R}$ is the countable union of open intervals. These open intervals are actually taken to be as large as possible:

Let $A\subseteq\mathbb{R}$ be open. Define an equivalence relation $\equiv$ on $A$ such that $x\equiv y$ if and only if, there is an open interval $(a,b)\subseteq A$ that contains both $x$ and $y$. That $\equiv$ is reflexive and symmetric is obvious.

It is transitive since the union of two open intervals that are not disjoint is again an open interval. This also shows that the equivalence classes are all open intervals. Since equivalence classes are disjoint, this shows that $A$ is the disjoint union of open intervals.

Now every open interval contains some rational number and disjoint intervals contain different rational numbers. It follows that $A$ is the disjoint union of countably many open intervals.

The only caveat is that the open intervals are not required to be finite. But a moments thought shows that at most two of the intervals are not finite, and we can write them as a countable union of finite non-disjoint intervals.

Note: The intervals $(x_n-\varepsilon_{n-} , x_n+\varepsilon_{n+})$ are actually the equivalence classes.

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Thanks for your reply, I'm sure that that way is just as good but to be consistent with my course I'd really like to understand what is meant by the above. –  Rosie May 20 '12 at 22:49
    
@Rosie: I've added an explanation for what your instructor does. –  Michael Greinecker May 20 '12 at 23:17
    
Thanks very much, it was the xn = yn bit which was confusing me, I thought they were different somehow. Great thanks –  Rosie May 20 '12 at 23:36
    
Sorry to go back to this, but looking over it a second time, what I dont understand is the very last part, how do we know that $\(x-\epsilon_x,x+\epsilon_x) ~~is~~ in (x_n-\epsilon_{n-},x_n+\epsilon_{n+})$, I would understand If it said that $\(x-\epsilon_x,x+\epsilon_x)~~ is~~ in (x -\epsilon_{n-},x+\epsilon_{n+})$ but I dont really understand the relationship between $\ x_n and x? –  Rosie Jun 7 '12 at 18:07
    
The point $x_n$ lies in $(x-\epsilon_x, x+\epsilon_x)$. The values $\epsilon_{n-}$ and $\epsilon_{n+1}$ are constructed so that $(x_n-\epsilon_{n-},x_n+\epsilon_{n+})$ is the largest open interva containing $x_n$. Since $(x-\epsilon_x, x+\epsilon_x)$ is an interval containing $x_n$, it must be a smaller such interval, a subset of $(x_n-\epsilon_{n-},x_n+\epsilon_{n+})$. –  Michael Greinecker Jun 7 '12 at 18:38

The steps to show that $\sigma(C_{\mathrm{open}}) \subset \sigma(C_0)$ are as follows:

Show that any open set $A$ can be written as a countable union of elements of $C_0$. From this it follows that $A \in \sigma(C_0)$. Since $A$ was 'any' set, this means that $C_{\mathrm{open}} \subset \sigma(C_0)$. From this it follows that $\sigma(C_{\mathrm{open}}) \subset \sigma(C_0)$.

To show that any open set $A$ can be written as a countable union of elements of $C_0$, I follow the basic construction above, but hopefully a little more clearly:

I will construct a countable union of intervals that are contained in $A$. Select $q \in A \cap \mathbb{Q}$, and define the interval $I_q$ as the largest open interval containing $q$ that is contained in $A$. The explicit construction is exactly the same as above:

Let $\overline{x}_q = \sup \{ x \; | \; [q, x) \subset A \}$, and $\underline{x}_q = \inf \{ x \; | \; (x,q] \subset A \}$. Then you can show that $I_q = (\underline{x}_, \overline{x}_q)$. It should be clear that $I_q \subset A$.

Let $B = \cup_{q \in \mathbb{Q} \cap A} I_q$, from above, it should be clear that $B \subset A$ (since each $I_q \subset A$). To finish, I just need to show that $A \subset B$.

Suppose $x\in A$, then since $A$ is open, we have $(x-\epsilon, x+\epsilon) \subset A$ for some $\epsilon > 0$. At least one $q \in \mathbb{Q}$ lies in this interval, and by construction $(x-\epsilon, x+\epsilon) \subset I_q$, hence we have $x \in B$. So we conclude that $A$ can be written as the countable union of intervals.

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