A continuous function on a closed interval takes every value between the two it takes at the ends. It is the first theorem in this course that says something a picture would suggest and that fails in the rationals, and that is exactly where completeness enters: the point at which the value is taken is produced by bisection, and bisection needs nested intervals.

A climber who is below the snow line at dawn and above it at dusk was at the snow line at some moment. The argument is not that the climber went slowly; it is that the height, followed continuously, cannot get past a level without touching it. On a mountain with a chasm at that level the conclusion fails, and the rationals are such a mountain.

Any level $y$ between the two end values is hit somewhere. The point $c$ is not computed from a formula: it is the common point of a nest of intervals, each half the last, chosen so that the level always lies between the two end values.

Let $f$ be continuous on $[a,b]$ and let $y$ lie between $f(a)$ and $f(b)$. Then for some . The theorem gives existence, not uniqueness, and says nothing about how many such points there are or where they lie. Its usual form is the case : a continuous function changing sign on an interval has a root in it.

• Continuity is essential: a step function skips every value between its two levels.
• The interval must be closed and bounded for the bisection to close on a point of the domain, and completeness is what makes it do so.
• It fails over $\mathbb{Q}$: $f(x)=x^2-2$ is continuous on the rationals in $[1,2]$, changes sign, and has no rational root.

Bisect, keeping the half where the sign change survives. At each stage the interval has the level below the value at one end and above it at the other, and one of its halves must have the same property, because the value at the midpoint is on one side of the level or the other. The intervals are closed, nested, and halve in length, so they close on a single point. Continuity at that point then forces its value to be the level: if it were strictly above or below, continuity would keep it on that side throughout a whole neighbourhood, which the shrinking intervals contradict.

Start with an interval on whose ends the level is straddled.

The value at the midpoint is on one side of the level, so one half inherits the straddle.

Closed nested intervals whose lengths shrink to nothing.

The nested interval property supplies one common point, and this is where completeness is spent.

Continuity at that point, in its sequential form, applied to both endpoint sequences.

Passing the inequalities to the limit traps the value from both sides.

The theorem promises a point and identifies none.

• A continuous function on a closed interval takes every value between its end values.
• The conclusion is existence, never uniqueness or location.
• Continuity and completeness are both essential, and the theorem fails over the rationals.
• The proof is bisection, and it is also an algorithm.

Great work! You now know how to:

• State the theorem with both of its hypotheses
• Use it to prove that a root exists
• Say where completeness is spent and why the rationals fail
• Keep existence apart from uniqueness and location