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Let A be the set defined as $A=\{x \in \mathbb{R}:0

$E_x =\{y \in \mathbb{R}:0

Then $ {\bigcap\limits_{x\in A} } E_x=\{\}$

Proof

$\forall y>0, y\notin\ E_x$ if $x

Hence $y\notin \bigcap\limits_{x\in A} E_x$

Now my problem with this is that if y>x then we are not operating with the set $E_x$ anymore, so how could this be a proof. What am I missing?

Thank you.

3 Answers 3

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So $A = (0,1]$ and $E_x =(0,x)$. The statement is an exercise in careful reading and translating it to a logical statement.

The claim is $\bigcap_{x \in A} E_x = \emptyset$.

By definition of the intersection, $p \in \bigcap_{x \in A} E_x$ iff

$$\forall x \in A: p \in E_x$$ the definition of $E_x$ then tell us:

$$\forall x \in A: 0 < p < x$$

by definition of $A$:

$$\forall x \in (0,1]: 0 < p

Because all $E_x \subseteq (0,1]$ our supposed $p$ lies in $(0,1]$ as well. Now we can take $x = \frac{p}{2} \in A$ in particular. But then the last statement becomes $0 < p < \frac{p}{2}$, which is false for all $p$.

This shows that no point $p\in \mathbb{R}$ lies in the intersection $\bigcap_{x \in A} E_x$.

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    Hm, why does A include zero as x had to be larger than zero. Not larger and equal to. – 2017-01-29
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    Yes, Henno it appears you've misread the definition of $A $. – 2017-01-29
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    @ALEXANDER yes, that is the more interesting case, I'll edit. – 2017-01-29
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You are right about the question. It just requires more careful phrasing, mostly what you have written is correct.

So let $y \in \mathbb R$, $y > 0$. Then, if $y > 1$ then $y \notin E_1$, so we are done. Otherwise, we know that $y>\frac{y}{2}>0$, so that $y \notin E_{\frac y2}$ (because of what you wrote in your post, taking $x = \frac{y}{2}$).Hence, $y$ is not in one of the $E_x$, hence it can't be in the intersection. So the intersection is empty.

It is good to have an idea of the answer, but rigorous writing and putting pieces together is a different matter. In this site, you should look out for answers which cover basic concepts rigorously, so that you can incorporate the same in your writing.

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    But if x=y/2 then y=2x. Hence we are not operating with a y$E_x$ – 2017-01-29
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    Note that $\bigcap\limits_{x \in A} E_x$ runs over ALL such $E_x$, not just any one PARTICULAR $x$. – 2017-01-29
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    See, we are checking if $y \in E_x$,right? If $y \in E_x$, then $0 < y < x$ must be true, right? However, in this case, if $x = \frac y2$, then $0 < y < \frac y2$ is *not true*, so $y \in E_x$ is *not true*, so $y \notin E_x$ is true in that case, which means it is not in the intersection of the $E_x$. @DavidWheeler The intersection may run over ALL such $E_x$, but if I show that $y$ does not belong to one such $E_x$, then certainly it will not belog to the **intersection** of the $E_x$ right? It may belong to the *union* , but the intersection demands membership in *every* given set. – 2017-01-29
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    Yes, that was my point. All we have to do is find one particular $E_{x'}$ that $y$ does not belong to, for any given $y$. It's rather similar to showing that $\bigcap\limits_{n = 1}^{\infty} (0,\frac{1}{n}) = \emptyset$, which uses a similar technique. – 2017-01-29
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    @DavidWheeler Ah, that's precisely what I did. But thank you for reaffirming that it was right anyway. – 2017-01-29
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Now my problem with this is that if y>x then we are not operating with the set $E_x$ anymore, so how could this be a proof. What am I missing?

That is not a problem; that is what you require.   If there is some $x$ in $A$, such that $y$ is not in that $E_x$, then $y$ is not in the intersection of all $E_x$ (over $x\in A$).   If that happens to be so for every real $y$ then the intersection is empty.

$y\in \bigcap_{x\in A} E_x$ is equivalent to $\forall x~\big((0

via:

$A=\{x\in \Bbb R: 0< x\leq 1\} = (0;1]$

$E_x=\{y\in \Bbb R:0

$\therefore~\bigcap_{x\in A} E_x = \{y\in\Bbb R: \forall x\in A:y\in E_x\}= \{y\in\Bbb R: \forall x\in (0;1]:y\in(0;x)\} =\{\}$