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I'm slightly unsure about how the theorem is presented to me in lecture...

The Brouwer fixed point theorem for $1$ dimension. Every continuous map $[0,1] \to [0,1]$ has at least one fixed point.

Well, the original theorem states the same except for that we deal with closed $n-$balls defined by

$$\bar{B}^n=\{(x_1,...,x_2) \in \mathbb{R}^n:x_1^2+...+x_n^2 \leq1\}$$

So, the definition of this ball should, in $1$ dimension, give me the closed interval $[-1,1]$ no? Since we square it and any reals that give values less than or equal to $1$ is in $\bar{B}^1$.

Doesn't this include the negative reals too? Why is my lecturer only considering the interval $[0,1]$ only? I suppose it works for $[0,1]$ if the theorem works for $[-1,1]$ but just curious as to why this is the case. I could ask my professor but he's away on holiday :/

Does anyone have any idea? Is this, say, a "classical" or "conventional" way to present this particular theorem or something?

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  • $\begingroup$ Any continuous function $f : [a,b] \to [a,b]$ has a fixed point. $\endgroup$ – Mambo May 3 '16 at 22:00
  • $\begingroup$ So if you're to answer my question, $[-1,1] \to [-1,1]$ has a fixed point too $\endgroup$ – John Trail May 3 '16 at 22:03
  • $\begingroup$ Intermediate Value theorem ! $\endgroup$ – Mambo May 3 '16 at 22:05
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Well, any two (bounded) closed intervals are homeomorphic to each other by way of stretching and translating.

The reason why in one dimension the theorem is stated that way is probably because the "unit interval" is more simplistic than the "unit ball". This is a matter of preference though.

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  • $\begingroup$ I see, so it is just basically about preference/simplicity. And by the way, when you say any two closed bounded sets are homeomorphic to each other, is it also regardless of dimensions? A set in $n$ dimension can be homeomorphic to another in $m$ dimension $n \neq m$ as long as both are closed and bounded? $\endgroup$ – John Trail May 3 '16 at 22:14
  • $\begingroup$ I say any two (bounded) closed intervals are homeomorphic to each other. In general for fixed $n$ any two closed balls in $\mathbb{R}^n$ (with the specific metric of this space) are homeomorphic. For different "dimensions" this is not true in general by virtue of making use of connectedness and point-removing arguments. $\endgroup$ – Alp Uzman May 3 '16 at 22:21
  • $\begingroup$ Alright, thanks a lot ! $\endgroup$ – John Trail May 3 '16 at 22:31
  • $\begingroup$ Not a problem! :) $\endgroup$ – Alp Uzman May 3 '16 at 22:36
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Any continuous mapping from a compact convex space to itself has at least one fixed point.

But if you don't want do just trust me on that (and why should you).

Suppose you have a mapping $f:[-1,1] \to [-1,1]$ that has a fixed point, $f(a) = a$.

and bijective $g:[-1,1] \to [0,1]$.

$(g\circ f\circ\ g^{-1})(x)$ has a fixed point at $g(a)$

Every continuous mapping from [-1,1] to itself has a fixed point.

Now you just have to find a g that fits the bill. e.g.

$g = 1/2(x+1)$

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