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Let $A = \lbrace (tr,1-t)\; | \; t \in [0,1], r \in \Bbb{Q}\rbrace$. Is it true that any continuous function from $A$ into $A$ has a fixed point?

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Is the topology induced from $\mathbb{R}^2?$ – Igor Rivin Jul 28 '13 at 20:59
Yes, the usual topology – dimo Jul 28 '13 at 21:01
I don't understand the problem. $A$ is homeomorphic to the interval $[0,1]$. So everything follows from the Brouwer fixed-point theorem (it's variant for the dimension 1). The statement of the problem looks strange. – Sergei Akbarov Jul 29 '13 at 11:24
@SergeiAkbarov Would you please tell why $A\cong [0,1]$ ? – dimo Aug 5 '13 at 9:01
This is a general theorem: every bijective and continuous map from a compact space to a Hausdorff space is a homeomorphism, see R.Engelking, General topology, 3.1.13, see also Wikipedia (the properties of Hausdorff spaces). – Sergei Akbarov Aug 5 '13 at 9:08
up vote 3 down vote accepted

Yes (I assume $A$ has the induced topology). The point $V:=(0,1)$, common endpoint of all segments $S_r:=\{(rt,1-t): t\in[0,1]\}$, is either fixed by the continuous function $f:A\to A$, or it is mapped into some $S_r\setminus\{V\}$. But then $f $ has a fixed point on $S_r$.

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To prove the last statement, we may consider another alternative: either $S_r$ is mapped into itself, so $f_{|S_r}$ has a fixed point there, or not, in which case some point $W\in S_r$ is mapped to $V$ by connection (this because $A\setminus\{V\}$ is disconnected): but then $f$ exchanges $V$ and $W$ and has a fixed point in between. – Pietro Majer Jul 28 '13 at 21:17
Either it does or not: if it does not, then it exchanges two points, and in this case you do not need $f(S_r)\subset S_r$ to conclude. – Pietro Majer Jul 28 '13 at 22:11
(sorry, "exchange" is not the right word. What I mean is: in the former case we have $V < f(V)$ and $W > f(W)$ in the order of $S_r$) – Pietro Majer Jul 29 '13 at 1:43

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