I have been pondering over this question for quite a long time. Please help. Is it possible to define some group operation on $[a,b]$ (with the usual topology inherited from $\mathbb R$) so that it becomes a topological group? Thanks for any help.
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No. Every continuous map $f$ from $[a,b]$ to itself has a fixed point, where $f(x) = x$. This even holds for all spaces $[a,b]^n$, $n \in \mathbb{N}$, by Brouwer's theorem, although the $n=1$ case is easy to see (consider $f(x) - x$ on $[a,b]$ and apply the intermediate value theorem between $a$ and $b$).
And no non-trivial topological group has this property: for every $a \neq e$, the map $f(x) = x \ast a$ has no fixed points.
Henno Brandsma
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also a nice method! – Hagen von Eitzen Oct 12 '14 at 21:12
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No. The points $a,b$ are "special" in $[a,b]$: They possess connected open neighbourhoods that remain connected after removal of the point. But if $[a,b]$ were a group, then "multiplication" with a suitable element transports $a$ to $\frac{a+b}2$, say, which is not special. This can't happen. (In other words: A topological group looks "the same" around each point)
Hagen von Eitzen
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