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Question: Let $U$ denote a subset of $\mathbb{R}$. We say that $U$ is an open set when for every $x \in U$ there is an $\epsilon > 0$ for which $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is contained in $U$. Show:

(i) For $a,b \in \mathbb{R}$ the interval $(a,b)$ is open, but the interval $[a,b]$ is not.

(ii) If $U,W \subset \mathbb{R}$ are open, so are $U \cup W$ and $U \cap W$.

(iii) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cup_{i \in I}U_{i}$ is also open.

(iv) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cap_{i \in I}U_{i}$ is not necessarily open.

My attempt: I think that with (i) i can conceptually understand everything:

$\forall x \in (a,b): a< x - \frac{|a-x|}{2} < x < x + \frac{|b-x|}{2} < b$

If $x> \frac{|b-a|}{2}$ let $\epsilon = \frac{|b-x|}{2}$

If $x< \frac{|b-a|}{2}$ let $\epsilon = \frac{|a-x|}{2}$

$\Rightarrow \forall x \in (a,b) \exists \epsilon > 0 : U_{\epsilon}(x) = (x - \epsilon, x +\epsilon) \Rightarrow$ the interval (a,b) is open.

and

max$[a,b] = b \Rightarrow \forall x \in [a,b]: b \geq x$

If $\epsilon > 0, b \ngeq b+ \epsilon \Rightarrow$ there exists an $x$ for which $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is not contained in $U \Rightarrow$ the interval is not open

With (ii) and (iii) i think i understand, but i am not sure how to start stating that they are true mathematically. to be sure, $\cup_{i \in I}U_{i}$ denotes several unions in a row, correct? i'm even reading a proof which goes:

Given $v \in U \cap W$, there exists an open ball $B$ of radius $r$ centered at $v$ contained in $U$, and there exists an open ball $B'$ of radius $r'$ centered at $v$ contained in $W$. Let $\delta =$ min$(r, r')$. Then the open ball of radius $\delta$ centered at $v$ is contained in $U \cap W$, which is therefore open.

Unfortunately i'm having trouble interpreting this and am wondering if there is another way to put it or approach it.

With (iv) i am not even sure intuitively why it should be true. i can't think of a situation where it wouldn't be true, i guess it has to do with whether the sets are finite or not, but i don't see how this would make a difference (especially given that openness is apparently preserved with infinite unions).

If anyone can help provide suggestions, corrections, hints, etc i appreciate it, thanks!

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1 Answer 1

up vote 4 down vote accepted

Intuitively, a set $U$ is open if every point in the set can be completely surrounded by points that are all in the set; here, "surrounded" means that an entire little interval centered at $x$ is completely contained in $U$, which means, in a sense, that $x$ is really inside $U$. A set is open if every point in the set is really "inside" the set. Compare with $[a,b]$ (which you show is not open), where one might say that although $b$ is in the set, it is "barely" in the set: move just a bit to the right, and you fall outside $[a,b]$.

That said, on to your work:

Your idea for the first part of (i) is okay, but the way you describe $\varepsilon$ is not. For example, say $x=1000$, $a=999.9$, and $b=1000.5$. Then $|b-a|/2 = .3$, and $x\gt |b-a|/2$, so you take $\varepsilon = 0.3$. But $(x-0.3,x+0.3)$ is not contained in $(a,b)$, because $x-0.3 = 999.7\lt a$.

What you really want to say is that $\varepsilon=\min(|x-b|/2, |x-a|/2)$.

For the second part, you've got it right, but there is no need to say "there exists an $x$ ...", just say that for every $\varepsilon\gt 0$, $(b-\varepsilon,b+\varepsilon)$ is not contained in $[a,b]$.

For (ii); not quite: $\cup_{i\in I} U_i$ is not "several unions in a row", because the set may not something that you can do "one after another"; maybe there is one for every real number! No, $\cup_{i\in I}U_i$ just means "the union of all the $U_i$"; this is the collection of all points that are in at least one of the $U_i$.

I think you mean $U\cap W$ when you said $V\cap V$. But you've go the idea right for the intersection. Conceptually: you know that you can find an open interval around $x$ that "works" for $U$, and one that works for $W$; to find one that works for both, just take the smaller one.

As for (iv), it's not true: you are being asked to show it is not true. Can you think of a way to get a closed interval as an intersection of open ones?

Added: Okay, here's a hint for (iv): for every positive integer $n$, $\frac{1}{n}$ is stricly larger than $0$; but the infinimum of $\{ \frac{1}{n}\mid n\in\mathbb{Z}_{\gt 0}\}$ is not strictly larger than $0$.

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thanks for the pointers, i edited the part where i had incorrectly written $V \cap V$. ok that makes a lot more sense when it comes to (i)! as for (iv) unfortunately i still can't think of a way to get a closed interval as an intersection of open ones.... –  ghshtalt Nov 25 '10 at 7:48
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@user3711: Think of the argument you give for finite intersections of open sets (you take the smallest $\epsilon$ among the available $\epsilon$'s). When and why will this argument fail for infinite intersections? Now use this fact to construct a counter example. –  user17762 Nov 25 '10 at 8:04

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