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I read it somewhere that the interval $(a,b]$ is a $G_{\delta}$ set as well as a $F_{\sigma}$ set Well I quickly wrote down

$(a,b]=\bigcap_{n=1}^{\infty} (a,b+\frac{1}{n})$

which makes it a $G_{\delta}$ set

Also

$(a,b]=\bigcup_{n=1}^{\infty} [a+\frac{1}{n},b]$

which would make it a $F_{\sigma}$ set.

My friend agrees with me on the first one. But strongly disagrees on the second one.

He says

$\bigcup_{n=1}^{\infty} [a+\frac{1}{n},b]=[a,b]$

I give him an argument that $a$ cannot belong to $\bigcup_{n=1}^{\infty} [a+\frac{1}{n},b]$ (let us denote this set by $I_n$ as for now) as if $a$ belongs to the union of $I_n$, then by Well Ordering Principle we'd get an $I_k$ such that $a\in I_k$.

That'd mean

$a\in [a+\frac{1}{k},b]$ for some $k\in \mathbb{N}$

Which would imply

$a+\frac{1}{k} \leq a$

That is

$\frac{1}{k} \leq 0$

which is a contradiction as $k$ being a natural number is positive (strictly) and so is its inverse.

At the end, he still disagrees. Am i wrong somewhere? How do i convince him (if I am correct, that is)

Note: He gives me an argument saying that $\lim_{n \to \infty} \frac{1}{n}=0$, so $a$ should belong to the set. I counter him by saying that it means that the value of $\frac{1}{n}$ tends to $0$ as $n$ increases and is never equal to it. He still isn't convinced! :/

Please clarify if I'm doing a blunder somewhere. Any help would be appreciated. Thanks.

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    $\begingroup$ Your argument looks good to me. $\endgroup$ – graydad Sep 22 '14 at 16:09
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    $\begingroup$ Your reference to the well ordering principle is unclear to me. This is simply the definition of the union. And certainly you mean $[a+\frac 1n,b]$ by $I_n$. $\endgroup$ – J.R. Sep 22 '14 at 16:11
  • $\begingroup$ Just look at the definition of union. $a$ is not in any of the sets $[a+1/n,b]$. $\endgroup$ – mathematician Sep 22 '14 at 16:12
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    $\begingroup$ it seems like he may getting confused because maybe he doesn't fully understand why the first example is correct. Maybe if you tell him first example has $(a,b]$ and not $(a,b)$ is because $b$ is included within each one of those intervals. With this the reason why your second example has answer $(a,b]$ and not $[a,b]$ is because $a$ is never inside any of those intervals $\endgroup$ – Kamster Sep 22 '14 at 16:19
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Here is a much easier way to see why $(a,b]$ is both $G_\delta$ and $F_\sigma$, granted you know the following two statements:

  1. If $A,B$ are both $F_\sigma$ ($G_\delta$) then $A\cap B$ and $F\cup B$ are $F_\sigma$ ($G_\delta$).
  2. Every open set is $F_\sigma$ (and every closed set is $G_\delta$).

Now simply note that:

$$(a,b]=(a,b+1)\cap[a,b]=(a,b-\delta)\cup[b-\delta,b]$$

(Where $\delta=\frac{b-a}3$, or some other sufficiently small number.)


In any case, you are right and he is wrong. $x$ belongs to a union if and only if it is in at least one of the sets unified. Since $a$ is in none of the closed intervals you unify, it cannot be in their union.

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You are exactly right, $a\in \bigcup_{n=1}^\infty [a+1/n,b]$ would imply that $a+1/n\le a\le b$ for some $n\in\{1,2,\dots\}$. This is clearly not possible by the argument that you have given.

Convince him by multiplying your last inequality by $k$. He will certainly agree that $1\le 0$ is not a true statement.

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In some sense, a union (particularly a nested union) is a limit of sets.

Many things we work with in mathematics are continuous, and we get in a habit of substituting

$$ \lim_{x \to a} f(x) = f(a) $$

without thinking about it.

This is what your friend is doing; he is equating in his mind

$$ \lim_{n \to \infty} \left[ a + \frac{1}{n}, b \right] = \left[ \lim_{n \to \infty} \left( a + \frac{1}{n} \right), b \right] $$

But in this sense, $[x,y]$ is simply not continuous in $x$ (or in $y$), so we can't do that with the limits. If you phrase things this way, it may help your friend realize where his error in thinking is, and consequently make him more receptive to seeing how the the rigorous definition of union shows that $a$ is not in your union.

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