An Interesting Application of Real Analysis in Geography

Does anybody here know that at any time instant there exist two antipodal points on the earth's equator that have the same temperature? The proof of this fact does not require any advanced knowledge in geography or meteorology. In fact, if we consider temperature to be a continuous function of space, then this fact can be derived from the principles of real analysis. First we shall prove the following proposition: "Let $f:[0,1]\rightarrow R$ be a continuous function such that $f(0)=f(1)$. Then $\exists c\in \left[ 0,\frac { 1 }{ 2 } \right]$ such that $f\left( c \right) =f\left( c+\frac { 1 }{ 2 } \right)$." Proving this proposition is very trivial. We simply consider the function $g\left( x \right) =f\left( x \right) -f\left( x+\frac { 1 }{ 2 } \right)$. Then $g(x)$ is continuous on $\left[ 0,\frac { 1 }{ 2 } \right]$ and we have $g\left( 0 \right) =f\left( 0 \right) -f\left( 0+\frac { 1 }{ 2 } \right) \\ =f\left( 0 \right) -f\left( \frac { 1 }{ 2 } \right) =f\left( 1 \right) -f\left( \frac { 1 }{ 2 } \right) \\ =-f\left( \frac { 1 }{ 2 } \right) +f\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) \\ =-\left\{ f\left( \frac { 1 }{ 2 } \right) -f\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) \right\} =-g\left( \frac { 1 }{ 2 } \right)$ Thus from location of roots theorem, $\exists c\in \left[ 0,\frac { 1 }{ 2 } \right]$ such that $g(c)=0$. This simply means that $f\left( c \right) =f\left( c+\frac { 1 }{ 2 } \right)$. Now we shall apply this to our given geographical proposition. Consider our equator to be the closed interval $[0,1]$ bent into a circle (this is a topological concept) with 0 and 1 as the same point on the equator. Then c and c+1/2 refer to two antipodal points. Temperature being a continuous function of the equator and temperatures at 0 and 1 being equal, we see the validity of our proposition.

For further reference to a curious reader let me tell that the generalized form of the above mathematical proposition is called Borsuk-Ulam theorem.