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Given a set:

$U = \{(x, y, z, w) : |x| < 1, |y| < 2, |z| < 3, |w| < 4\}$

We must formally (non-graphically, not that I'd ever be able to successfully graph such a set) prove that $U$ is open.

We did a similar, albeit simpler, problem in class which involved only $(x, y)$ and proving that the set of both of their absolute values was open. The route we took involved setting $\delta = \min(|1-x|,|1-y|,|-1-x|,|-1-y|)$ which made sense graphically. Although to be honest I'm not sure how the connection was made between that and a neighborhood of points in the set.

Consequently, I'm not sure how to approach a formal proof of this question. Any help will be appreciated.

Martin Sleziak
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michaelhankin
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  • I like to start my attack on this kind of proof with restating for myself a clear definition of the entity(ies) in question. In this case, what definition(s) do you have for open set? Sounds like you might be using a definition based on every point having a neighborhood around it contained in the set. So can you figure out a way to construct such a neighborhood given any point in $U$? – Todd Wilcox Apr 02 '15 at 02:23
  • The definition we were provided in class was if we can provide some $\delta$ s.t. for all points $p \in U, B_\delta (p) \subset U$ (where $B_\delta (p)$ is an open ball of radius $\delta$, centered at $p$) then the set $U$ is open. In this case I can't really come up with an applicable $\delta$, as I can't visualize the graph of the function. – michaelhankin Apr 02 '15 at 02:28

3 Answers3

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easiest way is that an intersection of open sets is open. So write it as the intersection of the following sets: $$ x < 1, x>-1, y < 2, y>-2, z< 3, z>-3, w>-4, w<4. $$ Proving that each of these is open is easy.

Mark Joshi
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Your set is a product of open intervals. Given a point $a$ in an open interval, you can find an open subinterval $(a-\delta,a+\delta)$ containing it. Do this for each coordinate, in each open interval, and then take the open ball of radius the minimum of the $\delta$s. This is a subset of the open hypercube made out of product of the subintervals, so it is also a subset of the original open set.

Chappers
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Given a point $p \in U$, I think you should first try to find an open four-dimensional box containing $p$ which is contained in $U$. Then you can find a small ball centred at $p$ which is contained in the box and hence in $U$.

Let $(a, b, c, d) \in U$, then $a \in (-1, 1)$, $b \in (-2, 2)$, $c \in (-3, 3)$, and $d \in (-4, 4)$. Let $\delta_1 = \min\{|1 - a|, |-1 - a|\}$; this is the shortest distance between $a$ and a point not in the interval $(-1, 1)$. If we leave the last three coordinates fixed, but we change the first coordinate $a$ by less that $\delta_1$, then the corresponding point remains in $U$. Likewise, we set

\begin{align*} \delta_2 &= \min\{|2 - b|, |-2 - b|\}\\ \delta_3 &= \min\{|3 - c|, |-3 - c|\}\\ \delta_4 &= \min\{|4 - d|, |-4 - d|\}. \end{align*}

Then $\delta_i$ measures how much we can change the $i^{\text{th}}$ coordinate and remain in $U$, provided we keep the other coordinates fixed. Now let $\delta = \min\{\delta_1, \delta_2, \delta_3, \delta_4\}$. Now if I vary any one of the coordinates by at most $\delta$, and keep the other coordinates fixed, then I remain in $U$. That is, $V = (a - \delta, a + \delta)\times(b - \delta, b + \delta)\times(c - \delta, c + \delta)\times(d - \delta, d + \delta)$ is contained in $U$. Furthermore, $V$ is open and contains $p$; this is the open four-dimensional box containing $p$ that I mentioned at the beginning.

Now you just need to find a ball centred at $p$ which is small enough to be contained in $V$ (and hence $U$). You can check that the ball $B(p, \delta)$ will do; intuitively, the change in any coordinate is less than or equal to the change in the distance.

Michael Albanese
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