Baby Rudin Chapter 4 Exercises 5 and 6: Solutions

The exercise numbers below refer to the commonly used third edition of Walter Rudin’s Principles of Mathematical Analysis. Exercise numbering can differ across editions or printings, so check that your copy has the same statements.

Exercise 5 asks you to prove that the set of zeros of a continuous function on $[a,b]$ is closed. Exercise 6 asks you to show that a continuous real-valued function on $[a,b]$ with no zeros must be either positive everywhere or negative everywhere.

Exercise 5: The zero set is closed

Let

$$S=\{x\in[a,b]:f(x)=0\},$$

where $f:[a,b]\to\mathbb{R}$ is continuous. To prove that $S$ is closed in $[a,b]$, use the sequence criterion for closed sets: show that whenever a sequence $(x_n)$ in $S$ converges to a point $x\in[a,b]$, that limit belongs to $S$.

For every $n$, membership in $S$ means $f(x_n)=0$. Since $f$ is continuous at $x$ and $x_n\to x$,

$$f(x)=\lim_{n\to\infty} f(x_n)=\lim_{n\to\infty}0=0.$$

Thus $x\in S$. Every convergent sequence in $S$ whose limit is in the domain has its limit in $S$, so $S$ is closed relative to $[a,b]$.

There is also a concise set-theoretic explanation. The singleton set $\{0\}$ is closed in $\mathbb{R}$, and $S=f^{-1}(\{0\})$. A continuous function has a closed preimage of every closed set, so $S$ is closed in its domain. The sequence proof makes the key role of continuity explicit.

Exercise 6: A continuous nonzero function has constant sign

Suppose $f:[a,b]\to\mathbb{R}$ is continuous and $f(x)\ne 0$ for every $x\in[a,b]$. We want to show that either $f(x)>0$ for every $x$, or $f(x)<0$ for every $x$.

Assume, toward a contradiction, that $f$ takes both positive and negative values. Then there are points $u,v\in[a,b]$ such that $f(u)>0$ and $f(v)<0$. If necessary, exchange their names so that $u<v$. The function is continuous on $[u,v]$, and $0$ lies strictly between $f(u)$ and $f(v)$. By the intermediate value theorem, some $c\in(u,v)$ satisfies $f(c)=0$. This contradicts the assumption that $f$ has no zeros.

Therefore $f$ cannot take both signs. Since it is nonzero at every point, all its values must be positive or all must be negative.

The interval condition matters: between any two points of an interval, every point between them is also in the domain. The same claim can fail on a disconnected domain. For example, define $f(x)=x$ on $[-1,0)\cup(0,1]$. This function is continuous on its domain and never zero there, but it takes both signs; the missing point $0$ prevents the intermediate value theorem from applying across the two pieces.

A worked example

Take $f(x)=x^2-1$ on $[0,2]$. Its zero set is

$$S=\{x\in[0,2]:x^2-1=0\}=\{1\},$$

which is closed in $[0,2]$. For any sequence of zeros $(x_n)$, every term must equal $1$, so its limit is also $1$, in agreement with the proof of Exercise 5.

For Exercise 6, instead take $g(x)=x^2+1$ on $[0,2]$. Since $g(x)\ge 1>0$, it has no zeros and is positive everywhere. In contrast, $h(x)=x-1$ takes both negative and positive values on $[0,2]$; the intermediate value theorem predicts a zero, and indeed $h(1)=0$.

Common mistakes

• Using continuity without an interval: The sign argument needs the intermediate value theorem on the segment between the two points. A disconnected domain may have a continuous, nonzero function with both signs.
• Forgetting relative closure: Exercise 5 proves that $S$ is closed as a subset of $[a,b]$. It does not, by itself, say that a zero set restricted to an arbitrary domain is closed in all of $\mathbb{R}$.
• Assuming the zero set is nonempty: Exercise 5 allows $S=\varnothing$. The empty set is closed, so no separate nonemptiness argument is needed.
• Claiming a zero set is always finite: It can contain infinitely many points, or even all of $[a,b]$ if $f$ is identically zero. Closedness is the conclusion, not finiteness.