Spivak saves the proof of the intermediate value theorem for his chapter on supremums and infemums.

His proof makes clear use of the completeness of reals, but why I struggle to see why it is dependent on completeness and not just continuity.

=================

1

On the rational numbers, the function 1×2>21_{x^2>2} is continuous.

– LutzL

2 days ago

1

Continuity can be used in many sets that are not the reals

– Zelos Malum

2 days ago

@LutzL ah, yea there it is. I was assuming that we were always considering the reals. Thanks

– Demetri P

2 days ago

=================

1 Answer

1

=================

f(x)=x2−2

f(x) = x^2 – 2

This function is continuous at every rational number xx.

f(0)=−2andf(2)=+2.

f(0) = -2 \quad \text{and} \quad f(2) = +2.

If the intermediate value theorem could be proved using only continuity, then we would conclude there is some rational number x0x_0 between 00 and 22 for which f(x0)=0f(x_0)=0. But there is none.