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Let $X$ be a connected topological space and $f:X\to \mathbb R$ be continuous. Further, we know that all $x\in X$ are local extrema. Does that imply that $f$ is constant?

I think in case $X$ is second-countable that should be the case because of the proof of Theorem 2 here (it is only stated for separable metric spaces there but I think the same proof works for second-countable spaces). But what about an arbitrary connected topological space $X$?

principal-ideal-domain
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1 Answers1

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The paper you link gives a counterexample already. Consider $X = [0,1]^2$ ordered lexicographically, i.e. $(x,y) < (x',y') \iff (x < x' \text{ or } (x = x' \text{ and y < y'}))$, and give $X$ the order topology. This is a connected space. Then the projection $p:(x,y) \mapsto x$ is a continuous function, and every point of $X$ is a local extremum. But $p$ is not constant.

Najib Idrissi
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  • @daw pointed that already out. But you a wrong in saying that every point is a local maximum. The point $(0,1)$ is no local maximum but a local minimum. – principal-ideal-domain Aug 13 '15 at 15:26
  • Why do you say that the projection is continuous? – Crostul Aug 13 '15 at 15:26
  • @Crostul The preimage of $(a,b)$ is $(a,b) \times [0,1]$, which is open in the order topology: if $a < x < b$, then let $a < u < x < v < b$, so that $(x,y)$ is in the "interval" between $(u,0)$ and $(v,1)$ (which is wholly in $(a,b) \times [0,1]$). – Najib Idrissi Aug 13 '15 at 15:27
  • @principal-ideal-domain He pointed this out in a comment, comments aren't for answers (I made this answer CW anyway). Thanks for the remark about maximum/minimum, I was a bit hasty here. – Najib Idrissi Aug 13 '15 at 15:30
  • @NajibIdrissi Can you find a counter-example where every point indeed is a local maximum? – principal-ideal-domain Aug 13 '15 at 15:44
  • @principal-ideal-domain This one isn't possible actually, see this question. (And now I feel silly when I stated that every point was a local maximum...) – Najib Idrissi Aug 13 '15 at 15:46