At every given point in time there are two points on the equator of the Earth that have the same temperature.

How do we know this? Well, here’s a proof. Let’s look at the equatorial plane which slices through the Earth at the equator. The equator is a circle which lies in that plane, and we can choose a coordinate system on the plane so that the point lies at the centre of the equator. For each point on the equatorial circle there is a point which lies diametrically opposite .

Points *x* and -*x*.

Now each point on the equator comes with a temperature . We can assume that the function , which allocates a temperature to each point, is continuous. That’s because temperature doesn’t suddenly jump up or down as you move around on the Earth.

Now consider the function

It is also continuous.

If this function is equal to for some point , then we are done because if

then

so the temperature at is the same as the temperature at .

If isn’t equal to anywhere, then let’s assume (without loss of generality) that there is a point at which so

This implies that

There is a result, called the *intermediate value theorem*, which says that if a continuous function is greater than at some point of its domain and less than at another, then it must equal at some point in between the two.

Illustration of the intermediate value theorem. If *t(x)>0* and *t(y)<0* and *t* is continuous, then there is a point *z* between *x* and *y* such that *t(z)=0*.

Thus, since and , there must be a point on the circle such that . So

which means that

So the temperature at the point is the same as the temperature at the point

The result actually holds for any circle on the Earth, not just the equator. In fact, the result is the one-dimensional case of the Borsuk-Ulam theorem, which says that for any continuous function from the circle to the real numbers there is a point such that

The more general version of the Borsuk-Ulam theorem says that for any continuous function from the -sphere to the set of -tuples of real numbers there is a point such that .