Define g:[0,1]→R by g(x)=f(x)−x. We know that g is continuous because f and the identity map both are, and by the limit laws. Applying the intermediate value theorem (problem #2) we see that there exists x∈[0,1] such that g(x)=0. But this means f(x)=x, so we are done.
Counterexample for the open interval: consider f:(0,1)→(0,1) defined by f(x)=x/2. First, we can verify that if 0<x<1 then 0<x/2<1/2<1, so f indeed maps to (0,1). To see that there is no fixed point, note that the only solution to x/2=x in R is 0, which is not in (0,1). (We can also view this graphically by plotting both y=x and y=x/2 and checking that they do not intersect in (0,1).)
EDIT: I’ve got another framing that I thought would be more useful for later problems, but I was wrong. I still think there is some value in understanding this proof as well.
In particular, look at this diagram on Wikipedia. It would be better if the whole upper triangle was blue and the whole lower triangle were red instead of just one side (you can arbitrarily decide whether to paint the rest of the diagonal blue or red). If x=0 and x=1 aren’t fixed points, then they must be blue and red respectively. If we split [0,1] into n components of size 1/n, then we can see where f(x) maps each such component to and form a line of colored points as in q1. Proving this using Sperner’s Lemma is then essentially the same as qu2.
Putting it right after #2 was highly suggestive—I wonder if this means there’s some very different route I would have thought of instead, absent the framing.
My solution for #3:
Define g:[0,1]→R by g(x)=f(x)−x. We know that g is continuous because f and the identity map both are, and by the limit laws. Applying the intermediate value theorem (problem #2) we see that there exists x∈[0,1] such that g(x)=0. But this means f(x)=x, so we are done.
Counterexample for the open interval: consider f:(0,1)→(0,1) defined by f(x)=x/2. First, we can verify that if 0<x<1 then 0<x/2<1/2<1, so f indeed maps to (0,1). To see that there is no fixed point, note that the only solution to x/2=x in R is 0, which is not in (0,1). (We can also view this graphically by plotting both y=x and y=x/2 and checking that they do not intersect in (0,1).)
EDIT: I’ve got another framing that I thought would be more useful for later problems, but I was wrong. I still think there is some value in understanding this proof as well.
In particular, look at this diagram on Wikipedia. It would be better if the whole upper triangle was blue and the whole lower triangle were red instead of just one side (you can arbitrarily decide whether to paint the rest of the diagonal blue or red). If x=0 and x=1 aren’t fixed points, then they must be blue and red respectively. If we split [0,1] into n components of size 1/n, then we can see where f(x) maps each such component to and form a line of colored points as in q1. Proving this using Sperner’s Lemma is then essentially the same as qu2.
Yeah, I did the same thing :)
Putting it right after #2 was highly suggestive—I wonder if this means there’s some very different route I would have thought of instead, absent the framing.