Find $\displaystyle A^c$ with respect to U=R in the following case. $\displaystyle A=(1, \infty) \cup (- \infty, -2]$

The answer is $\displaystyle ( - \infty , a] \cup [b, \infty)$ How do you know if it's an open interval or a closed interval?

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

The answer is $\displaystyle ( - \infty , a] \cup [b, \infty)$ How do you know if it's an open interval or a closed interval? - Dec 6th 2011, 11:08 PMAnnatalaRe: 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 $\displaystyle A^c$ must contain -2, but can't contain 1.

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

- Dec 6th 2011, 11:40 PMAnnatalaRe: open/closed intervals
- Dec 7th 2011, 02:39 AMemakarovRe: open/closed intervals
- Dec 7th 2011, 08:38 AMAnnatalaRe: open/closed intervals
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 $\displaystyle -\infty $ and $\displaystyle +\infty$), 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.