2

The temperature $T(x)$ at each point $x$ on the surface of Mars (a sphere) is a continuous function. Show that there is a point $x$ on the surface such that $T(x) = T(-x)$. Hint: Represent surface of Mars as $\{x \in \mathbb{R}^{3} : \|x\| = 1 \}$.

There is another post similar to this, but it only gives a hint and I didn't find the hint any more helpful. So here is my attempt:

Possible useful definition:

A path in $S \subset \mathbb{R}^{n}$ from $a$ to $b$, both points in $S$ is the image of a continuous function $\gamma$ from $[0,1]$ into $S$ such that $\gamma(0) = a$ and $\gamma(1) = b$.

Attempt:

let $f(x) = T(x) - T(-x)$.

What I envision myself wanting to do is using the existence of a path by somehow saying let $\gamma(0) = 0$ and $\gamma(1) = x$. My reasoning for this is that I somehow want to use the hint that $\|x\| = 1$. From here something should work out where $f(x) < 0$ and $f(x) > 0$ over some interval and apply IVT over this interval and I would get my equality.

Now trying to formalize this is where I am having trouble.....Is this the right idea?

D.C. the III
  • 5,719
  • 6
  • 53
  • 111

1 Answers1

3

No paths needed. Since the sphere is connected an $f$ is continuous, the image is $f$ is connected as well, i.e., an interval of the reals. As $f(-x)=-f(x)$, this interval contains $0$.

Michael Hoppe
  • 18,614
  • The main book we are learning from glosses over connectedness. It just mentions it very briefly at the end of the section. I've read and done work from other books showing connected sets get mapped to connected sets, but I don't think that is the approach they wanted me to take in this situation. I agree with your proof and the sleekness to, but is there a more albeit "clunky" way of deriving the solution? – D.C. the III Jun 25 '19 at 18:55
  • If you want to show that $f$ takes positive and negative values, I presume that it will boil down somehow to connectedness. – Michael Hoppe Jun 25 '19 at 19:09