# open/closed intervals

• December 6th 2011, 11:00 PM
Jskid
open/closed intervals
Find $A^c$ with respect to U=R in the following case. $A=(1, \infty) \cup (- \infty, -2]$

The answer is $( - \infty , a] \cup [b, \infty)$ How do you know if it's an open interval or a closed interval?
• December 6th 2011, 11:08 PM
Annatala
Re: open/closed intervals
Since open sets do not contain their boundary, their compliment must contain all points that are in the boundary of the original set (where open) and not contain the boundary when it's there (when closed).

Think of the interval as a set of points. Note that -2 is in A and 1 is not in A. This means $A^c$ must contain -2, but can't contain 1.

Your answer is not correct. The answer is $(-2, 1]$. -2 can't be in $A^c$ because it's in $A$. 1 must be in $A^c$ because it isn't in $A$. Is that clear? The compliment is just the set of points that aren't in its base.
• December 6th 2011, 11:25 PM
Jskid
Re: open/closed intervals
Yes that is clear, just one point is not. Why must it be an open interval for $+ \infty, - \infty$?
• December 6th 2011, 11:40 PM
Annatala
Re: open/closed intervals
Quote:

Originally Posted by Jskid
Yes that is clear, just one point is not. Why must it be an open interval for $+ \infty, - \infty$?

You were told that you were only dealing with real numbers, and $+ \infty$ and $-\infty$ are not real numbers. Therefore they are neither in A, nor its compliment.
• December 7th 2011, 02:39 AM
emakarov
Re: open/closed intervals
Quote:

Originally Posted by Jskid
Why must it be an open interval for $+ \infty, - \infty$?

What does "to be an open interval for" mean? I also assume you mean $(-\infty,+ \infty)$. Finally, this question is not related to the problem in the OP, is it?
• December 7th 2011, 08:38 AM
Annatala
Re: open/closed intervals
Quote:

Originally Posted by emakarov
What does "to be an open interval for" mean? I also assume you mean $(-\infty,+ \infty)$. Finally, this question is not related to the problem in the OP, is it?

They mean that infinity always appears as an open interval, I presume. Unless you're using the extended real line (which adds one extra point for each of $-\infty$ and $+\infty$), those values are not real numbers and as such they cannot be in the compliment.

Another way of saying that: the real line has no greatest or least element. It cannot be both closed and unbounded at the same time in the same direction, so "infinity" will never appear as the closed endpoint of an interval on the real numbers.