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I am thinking whether Every continuous function from $[0,1)$ on to $ [0,1)$ has a fixed point. I know about fixed point theorem for compact sets. The proof I came up fory mentioned problem is, if $f(0)=0$ then, we are done. If not, then it has to reach zero Somewhere in between and go to 1. So a compact subset of image is contained in the compact subset of domain or compact subset of domain is contained in the compact subset of image and thus it has a fixed point. Is it right??

Is it true that every interval mapped to itself with a continuous onto function, has a fixed point?

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If $f(0) \ne 0$ then $f(0) = c > 0$. There exists an $f(m) = 0; m > 0$

So let $g(x) = f(x) - x$ and....

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"Is it true that every interval mapped to itself with a continuous onto function, has a fixed point?"

Half-closed and closed, but not open intervals.

Consider $f(x) = x^2$ on $(0,1)$

This is informal and unclear, but I think if it as "requiring an anchor". If an interval is open you can slide or "drift" to the endpoints without reaching them but if the interval has a border point, that anchors one end of your leash and the rest stretches but can't drift.

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There is no general result to this effect. Consider $f: [0,1) \to [0,1)$ via: $$ f(x) = \frac{1}{2}x + \frac{1}{2}. $$ The issue here is that if I'm missing any point in the closure of $(0,1)$, I can shunt the fixed point off to a missing bit of the boundary to come off with a counterexample like the above.

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  • $\begingroup$ I said on to function. $\endgroup$ – jnyan Mar 12 '17 at 1:54
  • $\begingroup$ Ah, you might want to edit the question to say surjection more explicitly. $\endgroup$ – Pete Caradonna Mar 12 '17 at 16:41

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