Common Mistakes and Counterexamples

This page collects the wrong turns that come up most often in §1–§18. Each entry states a claim that sounds right, gives a short counterexample that breaks it, and says what is actually true. Citations such as Theorem 10.2 are links: hover over or focus one to see the statement.

A good habit: before using a theorem, check every hypothesis. Most of the counterexamples below work by dropping exactly one of them.

Writing proofs

An induction step that fails for the smallest case

Tempting. “In any group of $n$ people, everyone has the same birthday.” For $n = 1$ it is true. For the step, remove one person from a group of $n + 1$; the other $n$ share a birthday. Remove a different person; again the rest share a birthday. The two groups overlap, so all $n + 1$ do.

What goes wrong. Going from $n = 1$ to $n = 2$, the two groups of size $1$ do not overlap. The step must work for every $n$ from the basis on, including the first one. See §1.

Starting induction too early

Tempting. “$2^n > n^2$ for all $n$, by induction.”

Counterexample. It holds for $n = 1$, fails for $n = 2, 3, 4$, and holds for every $n \ge 5$. The step $2^n > n^2 \Rightarrow 2^{n+1} > (n+1)^2$ needs $2n^2 \ge (n+1)^2$, which is true only for $n \ge 3$, so the basis has to be $n = 5$. Induction may start at any integer, but its conclusion only covers what comes after the basis. See §1.

Letting N depend on n

Tempting. “Given $\varepsilon > 0$, take $N = n$.”

What is true. In Definition 7.1, $N$ may depend on $\varepsilon$ (and on the sequence), but it has to be chosen before $n$. The condition is then checked for every $n > N$. See §7.

Running the scratch work backwards

Tempting. “I started from $|s_n - s| < \varepsilon$ and simplified until I reached $n > \frac{1}{\varepsilon}$, so the proof is done.”

What is true. Working backwards is how you find $N$. The formal proof runs forwards, from $n > N$ to $|s_n - s| < \varepsilon$, and each step must be valid in that direction. Any assumption your estimate needs, such as “$n \ge 10$” or “the denominator is positive”, must be built into the choice of $N$. See §8.

Proving “for all” with a single example

Tempting. “$g(x_n) \to g(0)$ for the sequence $x_n = \frac1n$, so $g$ is continuous at $0$.”

What is true. One example can disprove a “for all” statement, but it cannot prove one. Continuity at $x_0$ needs $f(x_n) \to f(x_0)$ for every sequence $x_n \to x_0$ in the domain (Definition 17.1). For $g(x) = \sin\frac{\pi}{x}$ with $g(0) = 0$, the sequence $\frac1n$ gives values $0$, but $x_n = \frac{2}{4n + 1}$ gives values $1$. See §17.

Inequalities, suprema and infinity

Using the Triangle Inequality as a lower bound

Tempting. “$|a + b|$ is about the size of $|a| + |b|$.”

Counterexample. $|5 + (-5)| = 0$ while $|5| + |-5| = 10$. Theorem 3.7 only gives an upper bound. For a lower bound use $|a + b| \ge |a| - |b|$. See §3.

Subtracting or dividing inequalities

Tempting. “From $a < b$ and $c < d$, conclude $a - c < b - d$.”

Counterexample. $1 < 2$ and $0 < 5$, but $1 - 0 < 2 - 5$ is false. Inequalities pointing the same way can be added. Multiplying needs signs: multiplying by a negative number reverses the inequality (Theorem 3.2). See §3.

Assuming the supremum is the largest element

Tempting. “$\sup S$ is the biggest number in $S$.”

Counterexample. $\sup(0, 1) = 1$, and $1 \notin (0, 1)$. When $\sup S$ belongs to $S$ it is $\max S$; when it does not, $S$ has no maximum. See §4.

Keeping a strict inequality through a supremum or a limit

Tempting. “If $s < M$ for every $s \in S$, then $\sup S < M$.” Or: “If $s_n > 0$ for all $n$ and $s_n \to s$, then $s > 0$.”

Counterexample. Every element of $(0, 1)$ is less than $1$, and $\sup(0, 1) = 1$. Every $\frac1n$ is positive, and $\lim \frac1n = 0$. Suprema, limits and lim sups keep $\le$ but can turn $<$ into $=$. See §4, §9 and §12.

Mixing up the two halves of a sup proof

Tempting. “I showed $M$ is an upper bound of $S$, so $\sup S = M$.”

What is true. That proves only $\sup S \le M$. For $\sup S \ge M$ you also need elements of $S$ that come as close to $M$ as you like, or an argument that every upper bound is at least $M$. See §4.

Expecting rational sets to have rational suprema

Tempting. “A bounded set of rationals has a least upper bound, and it is rational.”

Counterexample. $S = \{r \in \mathbb{Q} : r^2 < 2\}$ is bounded above, but its least upper bound in $\mathbb{R}$ is $\sqrt2$, which is irrational (Corollary 2.3). Inside $\mathbb{Q}$ alone, $S$ has no least upper bound at all. The Completeness Axiom (Theorem 4.4) is exactly what $\mathbb{Q}$ lacks. See §2 and §4.

Doing arithmetic with ∞ − ∞

Tempting. “$\infty - \infty = 0$” or “$0 \cdot \infty = 0$.”

Counterexample. $n \to \infty$ and $n^2 \to \infty$, yet $n - n \to 0$, $n^2 - n \to +\infty$ and $n - n^2 \to -\infty$. Likewise $n \cdot \frac1n \to 1$ while $n^2 \cdot \frac1n \to +\infty$. These forms have no single value, which is why they are left undefined. See §5 and §9.

Limits of sequences

“The terms get closer and closer”

Tempting. “If $|s_n - s|$ decreases at every step, then $s_n \to s$.”

Counterexample. $\frac1n$ gets closer to $-1$ at every step, yet $\frac1n \to 0$. The distance must eventually drop below every $\varepsilon > 0$, not merely shrink. See §7.

Expecting convergence to be steady

Tempting. “A convergent sequence is eventually increasing or decreasing.”

Counterexample. $\frac{(-1)^n}{n} \to 0$, alternating sides of $0$ forever. Convergence says nothing about the path, only about where the terms end up. See §7.

Bounded, or monotone, is not enough

Tempting. “Bounded sequences converge.” Or: “Monotone sequences converge.”

Counterexample. $(-1)^n$ is bounded but diverges. $s_n = n$ is increasing but tends to $+\infty$. You need both together (Theorem 10.2). What is true for bounded sequences alone is that some subsequence converges (Theorem 11.5). See §10 and §11.

Splitting a limit before you know the pieces converge

Tempting. “$\lim(s_n + t_n) = \lim s_n + \lim t_n$, always.”

Counterexample. $s_n = (-1)^n$ and $t_n = -(-1)^n$: the sum is $0$ for every $n$, but neither limit on the right exists. Theorems 9.3 and 9.4 assume both limits exist as real numbers, and Theorem 9.6 also needs the limit in the denominator to be nonzero. See §9.

Solving for a limit that may not exist

Tempting. “$s_1 = 1$ and $s_{n+1} = 2s_n$. If $s = \lim s_n$, then $s = 2s$, so $s = 0$.”

Counterexample. $s_n = 2^{n-1} \to +\infty$. Passing to the limit in a recursion tells you what the limit must be if it exists. First prove it exists, for example with Theorem 10.2. See §10.

Small steps make a Cauchy sequence

Tempting. “If $|s_{n+1} - s_n| \to 0$, then $(s_n)$ is Cauchy.”

Counterexample. $\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \to 0$, yet $\sqrt n \to +\infty$. The Cauchy condition (Definition 10.8) controls $|s_m - s_n|$ for all pairs $m, n > N$, not just neighbours. The partial sums of $\sum \frac1n$ fail the same way. See §10.

Subsequences, lim sup and lim inf

Rearranging or repeating terms

Tempting. “$(s_2, s_1, s_4, s_3, \dots)$ is a subsequence of $(s_n)$.”

What is true. A subsequence uses indices $n_1 < n_2 < n_3 < \cdots$, so it keeps the original order and uses each term at most once. Neither swapping nor repeating terms is allowed (Definition 11.1). See §11.

One convergent subsequence

Tempting. “A subsequence converges to $s$, so the sequence does.”

Counterexample. $(-1)^n$ has the subsequence of even terms, which is constantly $1$, but $(-1)^n$ diverges. The true direction is the reverse: if the sequence converges, every subsequence converges to the same limit (Theorem 11.3). See §11.

Two subsequences with the same limit

Tempting. “Two subsequences converge to $s$, so the sequence converges to $s$.”

Counterexample. Let $s_n = 1$ when $n$ is a multiple of $3$ and $s_n = 0$ otherwise. The subsequences $(s_{3k+1})$ and $(s_{3k+2})$ are both constantly $0$, but $(s_n)$ diverges. The claim is true when the subsequences together use every index from some point on, for instance the even terms and the odd terms. See §11.

Confusing lim sup with sup

Tempting. “$\limsup s_n = \sup\{s_n : n \in \mathbb{N}\}$.”

Counterexample. For $s_n = \frac1n$, $\sup\{s_n\} = 1$ but $\limsup s_n = 0$. The supremum can be set by one early term; the lim sup ignores any finite number of terms (Definition 10.6). See §10 and §12.

Splitting lim sup over a sum

Tempting. “$\limsup(s_n + t_n) = \limsup s_n + \limsup t_n$.”

Counterexample. With $s_n = (-1)^n$ and $t_n = -(-1)^n$, the left side is $0$ and the right side is $1 + 1 = 2$. In general only $\le$ holds (when the right side makes sense). Products of limits need a convergent factor, as in Theorem 12.1. See §12.

Open, closed and compact sets

“Not open” means “closed”

Tempting. “Every set is either open or closed.”

Counterexample. $[0, 1)$ is neither. It is not open, because no interval around $0$ fits inside it. It is not closed, because its complement contains $1$ but no interval around $1$. $\mathbb{R}$ and $\varnothing$ are both open and closed. Closed means “the complement is open” (Definition 13.8), not “not open”. See §13.

Infinite intersections of open sets

Tempting. “An intersection of open sets is open.”

Counterexample. $\bigcap_{n=1}^\infty \left(-\frac1n, \frac1n\right) = \{0\}$, which is not open. Finite intersections of open sets are open. Dually, $\bigcup_{n=1}^\infty \left[\frac1n, 1\right] = (0, 1]$ is a union of closed sets that is not closed. See §13.

“Closed and bounded” as the definition of compact

Tempting. “A closed, bounded set is compact, in any metric space.”

Counterexample. In $\mathbb{Q}$ with the usual distance, $\{r \in \mathbb{Q} : 0 \le r \le 2\}$ is closed and bounded but not compact. The open sets $\{r \in \mathbb{Q} : |r - \sqrt2| > \frac1n\}$ cover it, and no finitely many of them do. The Heine–Borel Theorem 13.12 relies on the completeness of $\mathbb{R}^k$. See §13.

Series

Terms tending to 0

Tempting. “$a_n \to 0$, so $\sum a_n$ converges.”

Counterexample. $\sum \frac1n$ diverges although $\frac1n \to 0$. Corollary 14.5 runs the other way: if $a_n \not\to 0$, the series diverges. When $a_n \to 0$ you have learned nothing yet. See §14 and §15.

Confusing the series with its terms

Tempting. “$\sum_{n=1}^\infty \frac1{2^n}$ converges to $0$, because $\frac1{2^n} \to 0$.”

What is true. The sum of a series is the limit of its partial sums (Definition 14.2), not of its terms. Here the partial sums are $1 - \frac1{2^n}$, so the series converges to $1$. See §14.

Comparing in the wrong direction

Tempting. “$\frac1{n^2} \le \frac1n$ and $\sum \frac1n$ diverges, so $\sum \frac1{n^2}$ diverges.”

What is true. To prove convergence, bound the terms above by a convergent series. To prove divergence, bound them below by a divergent one. The other two combinations prove nothing. See §14.

Comparing with signed terms

Tempting. “$b_n \le a_n$ and $\sum a_n$ converges, so $\sum b_n$ converges.”

Counterexample. $b_n = -1 \le \frac1{n^2} = a_n$, and $\sum b_n = -\infty$. The Comparison Test (Theorem 14.6) needs $|b_n| \le a_n$, not just $b_n \le a_n$. See §14.

Ratios less than 1, or tending to 1

Tempting. “$\left|\frac{a_{n+1}}{a_n}\right| < 1$ for every $n$, so $\sum a_n$ converges.” Or: “the ratios tend to $1$, so the series diverges.”

Counterexample. For $\sum \frac1n$ every ratio $\frac{n}{n+1}$ is below $1$, yet the series diverges. And the ratios tend to $1$ both for $\sum \frac1n$, which diverges, and for $\sum \frac1{n^2}$, which converges. The Ratio Test (Theorem 14.8) needs the ratios to stay below a fixed $r < 1$, and it says nothing when $\limsup$ of the ratios is $1$. The same applies to the Root Test (Theorem 14.9). See §14.

Convergent means absolutely convergent

Tempting. “If $\sum a_n$ converges, so does $\sum |a_n|$.”

Counterexample. $1 - \frac12 + \frac13 - \frac14 + \cdots$ converges by the Alternating Series Theorem 15.3, but $\sum \frac1n$ diverges. Only the reverse is true: absolutely convergent series converge (Corollary 14.7). See §14 and §15.

Alternating signs are enough

Tempting. “The signs alternate and the terms tend to $0$, so the series converges.”

Counterexample. $1 - 1 + \frac12 - \frac14 + \frac13 - \frac19 + \cdots$, whose $k$-th pair is $\frac1k - \frac1{k^2}$. The partial sums at the end of each pair are $\sum_{j \le k} \left(\frac1j - \frac1{j^2}\right) \to +\infty$. Theorem 15.3 needs the absolute values of the terms to decrease. See §15.

Using the integral test without “decreasing”

Tempting. “$\sum f(n)$ and $\int_1^\infty f(x)\,dx$ always behave alike.”

Counterexample. $f(x) = \sin^2(\pi x)$ is $0$ at every integer, so $\sum f(n) = 0$, yet the integral is $+\infty$. The comparison between sums and integrals in §15 relies on $f$ being decreasing (and positive). See §15.

Rearranging a conditionally convergent series

Tempting. “Changing the order of the terms cannot change the sum.”

Counterexample. Take the terms of $1 - \frac12 + \frac13 - \frac14 + \cdots$, which sums to $\log 2$, and list them as one positive term followed by two negative ones. The rearranged series sums to $\frac12 \log 2$. Rearranging is safe only for absolutely convergent series. See §15.

Treating 0.999… as slightly less than 1

Tempting. “$0.999\ldots$ is the closest number below $1$.”

What is true. $0.999\ldots$ is the limit of $0.9, 0.99, 0.999, \ldots$, and that limit is exactly $1$. Two different expansions name the same number, and this happens only for numbers of the form $\frac{m}{10^k}$ (Theorem 16.3). There is no “closest number below $1$”: between any two reals there is another. See §16.

Continuity

“Continuous” means “the graph has no gaps”

Tempting. “$f(x) = \frac1x$ is discontinuous at $0$.”

What is true. Continuity is defined only at points of the domain (Definition 17.1), and $0$ is not in the domain of $\frac1x$. So $\frac1x$ is continuous: at every point where it is defined. Whether it could be extended continuously to $0$ is a question about limits of functions. See §17.

Continuous at a point means continuous nearby

Tempting. “If $f$ is continuous at $x_0$, it is continuous on some interval around $x_0$.”

Counterexample. Let $f(x) = x$ for rational $x$ and $f(x) = 0$ for irrational $x$. Since $|f(x)| \le |x|$, $f$ is continuous at $0$. At any $a \ne 0$, rationals tending to $a$ give values tending to $a$ and irrationals give $0$, so $f$ is continuous only at $0$. See §17.

Sums of discontinuous functions are discontinuous

Tempting. “If $f + g$ is continuous, then $f$ and $g$ are.”

Counterexample. Let $f = 1$ on $\mathbb{Q}$ and $0$ elsewhere, and $g = 1 - f$. Both are discontinuous at every point, but $f + g = 1$. Theorem 17.4 goes one way only: continuous parts give a continuous sum, product and quotient. See §17.

The Extreme Value Theorem on the wrong kind of set

Tempting. “A continuous function on an interval has a maximum and a minimum.”

Counterexample. $f(x) = x$ on $(0, 1)$ is bounded but has no maximum or minimum. $\frac1x$ on $(0, 1]$ is unbounded. $f(x) = x$ on $[0, \infty)$ is unbounded. Theorem 18.1 needs a closed, bounded interval $[a, b]$. See §18.

The Intermediate Value Theorem without an interval, or over ℚ

Tempting. “$f$ is continuous, negative somewhere and positive somewhere, so $f$ has a zero.”

Counterexample. $\frac1x$ on $[-1, 0) \cup (0, 1]$ is continuous, takes the values $-1$ and $1$, and is never $0$: the domain is not an interval. And on $\mathbb{Q}$, $x^2 - 2$ is negative at $1$ and positive at $2$ but has no rational zero: the proof of Theorem 18.2 needs completeness. Even when the theorem applies, it gives existence only, not where the zero is or how many there are. See §18.

The intermediate value property implies continuity

Tempting. “If $f$ takes every value between any two of its values, then $f$ is continuous.”

Counterexample. $f(x) = \sin\frac1x$ for $x \ne 0$ and $f(0) = 0$. Every interval around $0$ contains points where $f$ takes every value in $[-1, 1]$, so $f$ has the intermediate value property, but $f$ is not continuous at $0$. Exercise 18.12 works through this example. For strictly increasing functions the converse does hold (Theorem 18.5). See §18.