Fσ and Gδ sets in the Euclidean Topology

**magus** · January 1st 2011, 02:56 AM

I have a four part question to which I can only seem to answer a few parts.

(i) Prove that each open interval is a $F_\sigma$ set in $\mathbb{R}$ and each closed interval as well.

(ii) Prove that each open and closed interval is an $G_\delta$ set.

(iii) Prove that $\mathbb{Q}$ is an $F_\sigma$ set.

(iv) Prove that the compliment of an $F_\sigma$ set is an $G_\delta$ set and vice versa.

In order to prove (i) I thought of a union of closed intervals that expanded but were limited between a and b. Since all real numbers have a sequence with itself as a limit, Cauchy, I thought about taking the monotonically decreasing subsequence and monotonically increasing subsequence of each of them to form an expanded interval. For instance:

There exists and monotonically increasing sequence $b_n$ and a monotonically decreasing sequence $a_n$ .

and so

$\bigcup _{n=1}^{\infty}[a_n,b_n]=(a,b)$

To prove that [a,b] is the union of a closed interval just make take a cousin cover of that interval.

For (ii) I've got no idea how to make the open interval but the closed interval is easy as it can be the intersection of two overlapping open intervals.

On (iii) and (iv) I've got no clue.

Could someone tell me if what I have done is right and give me a few hints as to how to solve the remaining portions?

**magus** · January 1st 2011, 04:41 AM

I just got the answer of (iv) and it's obvious now to me. I just needed to remember that $(A\cup B)'=A'\cap B'$

I'm wondering if the answer to (iii) is this: take all the singltons of the rational numbers and (since they are countable and closed) take the union of them and that's the $F_\sigma$ set.

**Plato** · January 1st 2011, 04:49 AM

If $[a,b]$ is a closed interval in $\mathbb{R}$ define $\ell = \frac{{b - a}}{2}$ .
Now define $F_n=\left[ {a + \frac{\ell }{{2^n }},b - \frac{\ell }{{2^n }}} \right]$ and $G_n = \left( { - \infty ,b + \frac{\ell }{{2^n }}} \right) \cap \left( {a - \frac{\ell }{{2^n }},\infty } \right)$ .
Use those to do #1 . Note that using this approach we have a definite set of intervals. Also note that if is only proves #1 for a finite closed interval.

What about $[a,\infty)~?$

**magus** · January 1st 2011, 07:28 AM

I can see it for $F_n$ but the unions of the $G_n$ give $\mathbb{R}$ with some closed interval removed. Also it's constructed with open sets and an $F_\sigma$ set has to be made with closed sets.

As for $(a,\infty)$ , $(-\infty,a)$ and $(-\infty,a)\cup(b,\infty)$

$\bigcup _{n=1}^{\infty}[b+\frac{1}{2^n},n!]$
$\bigcup _{n=1}^\infty[-n!,a-\frac{1}{2^n}]$
and the third can just be the union of the unions of closed sets we made for the those two.
$\bigcup _{n=1}^\infty[n!,a-\frac{1}{2^n}]\cup \bigcup _{n=1}^{\infty}[b+\frac{1}{2^n},n!]$

Is this at all right?

**Plato** · January 1st 2011, 07:40 AM

No, the point is that $\left[ {a,b} \right] = \bigcap\limits_n {G_n }$ which shows that a bounded closed interval is $G_{\delta}$ .
$\left[ {a,\infty } \right) = \bigcap\limits_n {\left( {a - \frac{1} {n},\infty } \right)}$ shows it for unbounded closed interval.

You approach to $\mathbb{Q}$ being $F_{\sigma}$ is correct.

**magus** · January 1st 2011, 12:35 PM

Wouldn't the successive intersections of $G_n$ leave a hole in the center?

**Plato** · January 1st 2011, 12:40 PM

Originally Posted by magus

Wouldn't the successive intersections of $G_n$ leave a hole in the center?

$\left[ {a,b} \right] \subseteq \left( { - \infty ,b + \frac{\ell } {{2^n }}} \right) \cap \left( {a - \frac{\ell } {{2^n }},\infty } \right) = \left( {a - \frac{\ell } {{2^n }},b + \frac{\ell } {{2^n }}} \right)$

**magus** · January 1st 2011, 01:13 PM

Oops sorry I read it wrong. Thank you for all the help.

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