1.1 Intervals
Let \(a<b\) be real numbers. There are four bounded intervals with end points \(a\) and \(b\): \[[a,b],\qquad (a,b),\qquad [a,b),\qquad (a,b].\] For each real number \(c\) there are four unbounded intervals, or half-lines: \[\{x \in \mathbb {R}: x \leq c\},\qquad \{x\in \mathbb {R}: x < c\},\qquad \{x\in \mathbb {R}: x\geq c\},\qquad \{x\in \mathbb {R}: x > c\}.\] The ninth interval is the whole real line \(\mathbb {R}\).
Note.
- 1.
- In interval notation the half-lines are \[(-\infty , c],\qquad (-\infty , c),\qquad [c,\infty ),\qquad (c,\infty ),\] and \(\mathbb {R} = (-\infty , \infty )\). The symbol \(\infty \) (read “infinity”) is not a real number; it records only that the interval has no end on that side, since \(-\infty < x < \infty \) for every real \(x\).
- 2.
- For \(a \in \mathbb {R}\) we also allow the degenerate cases \((a,a) = [a,a) = (a,a] = \emptyset \) and \([a,a] = \{a\}\), a singleton set. These also count as intervals.
Example 1.1.1. Write each set as an interval or a union of intervals: (a) \(\{x\in \mathbb {R}: |x-3|<2\}\); (b) \(\{x\in \mathbb {R}: x^2\leq 4\}\); (c) \(\{x\in \mathbb {R}: x^2-5x+6<0\}\); (d) \(\{x\in \mathbb {R}: |2x+1|\geq 3\}\).
Solution. (a) \(|x-3|<2\) means \(-2<x-3<2\), that is, \(1<x<5\). The set is \((1,5)\).
(b) \(x^2\leq 4\) means \(-2\leq x\leq 2\). The set is \([-2,2]\).
(c) \(x^2-5x+6 = (x-2)(x-3)\) is negative exactly when the factors have opposite signs, that is, when \(2<x<3\). The set is \((2,3)\).
(d) \(|2x+1|\geq 3\) means \(2x+1\geq 3\) or \(2x+1\leq -3\), that is, \(x\geq 1\) or \(x\leq -2\). The set is \((-\infty ,-2]\cup [1,\infty )\), which is not an interval: it contains \(-2\) and \(1\) but not \(0\).
Example 1.1.2. Show that the intersection of two intervals is an interval, but their union need not be.
Solution. A set \(I\) is an interval exactly when \(x,y\in I\) and \(x<z<y\) imply \(z\in I\). If \(I\) and \(J\) have this property and \(x,y\in I\cap J\) with \(x<z<y\), then \(z\in I\) and \(z\in J\), so \(z\in I\cap J\). For instance \([0,3)\cap (1,5] = (1,3)\).
For unions, \([0,1]\cup [2,3]\) contains \(1\) and \(2\) but not \(\frac {3}{2}\), so it is not an interval. The union of two intervals is an interval when they overlap or touch, as in \([0,1)\cup [1,2] = [0,2]\).
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.