Each challenge below states a claim that sounds plausible but is false . Pick the candidate that is a real counterexample. A wrong pick tells you why it fails, which is often as instructive as the answer. The full list of pitfalls is on Common mistakes .
Bounded means convergent?
Claim Every bounded sequence converges.
Which of these breaks it?
s n = 1 n s_n = \frac1n It is bounded, but it converges to 0 0 s sub n equals 1 over n s n = ( − 1 ) n s_n = (-1)^n Bounded by 1 1 − 1 -1 1 1 s sub n equals negative 1, raised to the n s n = n s_n = n Not bounded, so the claim says nothing about it. s sub n equals n What is true. A bounded sequence need not converge, but a bounded
monotone sequence does (
Theorem 10.2 ), and every bounded sequence has a convergent subsequence (
Theorem 11.5 ).
Terms shrinking is not enough
Claim If s n → 0 s_n \to 0 ∑ s n \sum s_n
Which of these breaks it?
s n = 1 2 n s_n = \frac{1}{2^n} The series is geometric with ratio 1 2 \frac12 s sub n equals 1 over 2 to the n s n = 1 n s_n = \frac1n The terms tend to 0 0 s sub n equals 1 over n s n = 1 n 2 s_n = \frac{1}{n^2} This series converges (compare with p = 2 p = 2 s sub n equals 1 over n squared What is true. s n → 0 s_n \to 0 is
necessary for
∑ s n \sum s_n to converge, not sufficient.
Sums and products of divergent sequences
Claim If s n s_n t n t_n s n + t n s_n + t_n
Which of these breaks it?
s n = n s_n = n t n = n t_n = n The sum is 2 n 2n s sub n equals n; t sub n equals n s n = ( − 1 ) n s_n = (-1)^n t n = ( − 1 ) n + 1 t_n = (-1)^{n+1} Each diverges, but s n + t n = 0 s_n + t_n = 0 n n s sub n equals negative 1, raised to the n; t sub n equals negative 1, raised to the n plus 1 s n = 1 n s_n = \frac1n t n = 2 t_n = 2 Both of these converge, so they are not a pair of divergent sequences. s sub n equals 1 over n; t sub n equals 2 What is true. The limit laws need both sequences to converge . If one converges and the other diverges, the sum does diverge.
Is the supremum in the set?
Claim Every nonempty bounded set of reals has a largest element.
Which of these breaks it?
[ 0 , 1 ] [0, 1] It contains its supremum 1 1 max = 1 \max = 1 closed interval from 0 to 1 ( 0 , 1 ) (0, 1) Bounded, but for any x < 1 x < 1 x + 1 2 \frac{x+1}{2} open interval from 0 to 1 { 1 , 2 , 3 } \{1, 2, 3\} A finite set always has a maximum. the set 1, 2, 3 What is true. A bounded set has a supremum (completeness), which need not belong to the set.
Continuous implies uniformly continuous?
Claim Every continuous function on an interval is uniformly continuous.
Which of these breaks it?
f ( x ) = x 2 f(x) = x^2 [ 0 , 1 ] [0, 1] Continuous on a closed bounded interval, so it is uniformly continuous. f of x equals x squared on the closed interval from 0 to 1 f ( x ) = 1 x f(x) = \frac1x ( 0 , 1 ) (0, 1) Continuous, but near 0 0 δ \delta ε \varepsilon f of x equals 1 over x on the open interval from 0 to 1 f ( x ) = 3 x + 1 f(x) = 3x + 1 R \mathbb{R} Take δ = ε / 3 \delta = \varepsilon/3 f of x equals 3 x plus 1 on the reals What is true. Continuous on a closed bounded interval implies uniformly continuous. Drop either word and it can fail.
Intermediate values without continuity
Claim If f f [ 0 , 1 ] [0, 1] f ( 0 ) < 0 < f ( 1 ) f(0) < 0 < f(1) f f
Which of these breaks it?
f ( x ) = x − 1 2 f(x) = x - \frac12 Continuous, and it has a zero at 1 2 \frac12 f of x equals x minus one half f ( x ) = − 1 f(x) = -1 x < 1 2 x < \frac12 f ( x ) = 1 f(x) = 1 x ≥ 1 2 x \ge \frac12 Negative at 0 0 1 1 0 0 f of x equals negative 1 for x less than one half, and f of x equals 1 for x greater than or equal to one half f ( x ) = 2 x − 1 f(x) = 2x - 1 Continuous, so the Intermediate Value Theorem applies and gives a zero. f of x equals 2 x minus 1 What is true. The Intermediate Value Theorem needs
f f to be
continuous on
[ 0 , 1 ] [0, 1]