Continuity and Existence

A thermostat reading can pass from below a target to above it. If temperature varies continuously, it must equal the target somewhere between the readings. This conclusion concerns an unobserved instant. It follows from a mathematical assumption about the entire interval, not from the two readings alone. A sensor that jumps between quantized readings need not display every intermediate number.

Matching nearby behavior to the actual value

A function is continuous at $a$ when $f(a)$ exists, the finite limit $\lim_{x\to a}f(x)$ exists, and that limit equals $f(a)$. For an interior point these are two-sided conditions. At an included endpoint of an interval, use the appropriate one-sided limit. Continuity on $[a,b]$ means continuity at all interior points, right continuity at $a$, and left continuity at $b$.

For $f(x)=(x^2-1)/(x-1)$, the limit at $1$ is $2$ but the function is undefined there. Defining $f(1)=2$ creates a continuous extension. Defining $f(1)=7$ leaves a removable discontinuity: changing one value can repair it. A jump with different left and right limits cannot be repaired that way. Neither can the unbounded behavior of $1/x$ at zero.

Polynomials are continuous everywhere on the real line. Rational functions are continuous wherever their denominators are nonzero. Compositions of continuous functions are continuous where the composition is defined. These are consequences of limit laws. They let us verify continuity without repeating a tolerance proof at every point.

Joining two formulas

Let $f(x)=kx+1$ for $x<2$ and $f(x)=x^2-1$ for $x\geq2$. Each piece is continuous on its own interval. At the join, the left limit is $2k+1$, the right limit is $3$, and $f(2)=3$. Continuity requires $2k+1=3$, hence $k=1$. Checking only that both formulas are polynomials misses the join, which is the only point that needs extra work.

The intermediate value theorem

If $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then some $c\in[a,b]$ satisfies $f(c)=N$. If $N$ lies strictly between the endpoint values, such a $c$ lies in $(a,b)$. The theorem guarantees existence. It does not guarantee one solution, a formula for it, or that the function is monotone.

For $p(x)=x^3+x-1$, continuity holds on $[0,1]$, and $p(0)=-1$, $p(1)=1$. Therefore a zero lies in $(0,1)$. It is unique because for $v>u$,

$$p(v)-p(u)=(v-u)(v^2+uv+u^2+1)>0.$$

The second factor is positive since $v^2+uv+u^2=(v+u/2)^2+3u^2/4\geq0$. We supplied a separate monotonicity argument; IVT alone would not establish uniqueness.

The continuity hypothesis is essential. A function equal to $-1$ for $x<0$ and $1$ for $x\geq0$ changes sign across $[-1,1]$ without ever equalling zero. A sign change is not a substitute for continuity.

Bisection with an error bound

If a continuous function has opposite signs at the endpoints, evaluate its midpoint. Keep the half whose endpoints still bracket a sign change, unless the midpoint is already a root. Repeating gives nested intervals. After $n$ halvings, the bracket length is $(b-a)/2^n$, and its midpoint is within $(b-a)/2^{n+1}$ of at least one root remaining in the bracket.

For $p$ above, $p(0.5)=-0.375$, while $p(0.75)=0.171875$. A root therefore lies in $(0.5,0.75)$. The midpoint $0.625$ has a guaranteed error at most $0.125$ at this stage. To guarantee midpoint error no greater than $0.001$ from the initial unit interval, nine halvings suffice because $1/2^{10}<0.001$. A small function residual alone is a different notion of error; a very flat function can have a small residual far from a root.

The extreme value theorem

If $f$ is continuous on a closed bounded interval $[a,b]$, it attains an absolute maximum and an absolute minimum on that interval. The conclusion is stronger than having a finite upper or lower bound: actual inputs achieve the extreme outputs.

Both interval conditions matter. On $(0,1)$, the continuous function $f(x)=x$ attains neither its supremum $1$ nor its infimum $0$. On $\mathbb R$, the continuous function $x^2$ has no absolute maximum. On $[0,1]$, a discontinuous function can be unbounded: set $f(0)=0$ and $f(x)=1/x$ for $x>0$.

IVT and EVT are foundational real-number theorems. Their full proofs depend on completeness of the real numbers. Here we use their stated conclusions and identify the hypotheses each application needs. Later, EVT will ensure optimization problems have candidates worth seeking.

Exercises

1. Find $k$ making $f(x)=kx+2$ for $x<1$, $f(x)=x^2+4$ for $x\geq1$ continuous.
2. Prove that $x^3-2$ has a zero in $(1,2)$. Is uniqueness supplied by IVT?
3. How many bisection halvings reduce an initial bracket of length $4$ to midpoint error at most $0.01$?
4. Give a continuous function on an open bounded interval with no maximum.
5. Does EVT imply that a maximum occurs at an interior point?
6. For $f(x)=1/x$ on $[-1,1]\setminus\{0\}$, explain why opposite endpoint signs do not establish a zero.

Worked solutions

1. The right value and limit at $1$ are $5$; $k+2=5$ gives $k=3$.
2. The polynomial is continuous and has endpoint values $-1$ and $6$. IVT establishes a root; uniqueness follows separately from strict increase of the cube function.
3. Require $4/2^{n+1}\leq0.01$. Eight halvings give $4/512=0.0078125$; seven give $0.015625$, so eight suffice.
4. $f(x)=x$ on $(0,1)$ has supremum $1$ that is never attained.
5. No. For $f(x)=x$ on $[0,1]$, the maximum is at $1$.
6. The function is not continuous on the full closed interval; it is undefined at zero. IVT does not apply.