AlanH
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 Feb6 awarded Nice Question Nov25 awarded Popular Question Nov21 awarded Popular Question Oct6 awarded Yearling Jul16 awarded Civic Duty Jun28 awarded Taxonomist Jun26 accepted Showing path connected matrices of a group $G$ is a normal subgroup Jun26 comment Showing path connected matrices of a group $G$ is a normal subgroup In the first part, to allow for the inverse of $B$, do I need $\tau^{-1}$ to be continuous too or is it enough to say $\tau$ is continuous? Is this equivalent to saying if $B\in H$, we want to show $B^{-1}$ is in $H$? And you don't need to edit the post just to fix $I$, it's clear now. Jun26 asked Showing path connected matrices of a group $G$ is a normal subgroup Jun21 revised Showing the topologists sine curve is connected (slight variation) added 70 characters in body Jun21 asked Showing the topologists sine curve is connected (slight variation) Jun19 accepted Understanding equivalent definitions of left cosets Jun19 revised Understanding equivalent definitions of left cosets deleted 1 characters in body Jun19 asked Understanding equivalent definitions of left cosets Jun18 accepted Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$ Jun17 comment Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$ So the unit intveral on the x-axis wouldn't be open in $\Bbb{R}^2$ right? Jun17 comment Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$ @Berci Thanks for that Jun17 comment Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$ Yes, but it is still a union of open balls. So in some sense, are open balls the basis of all open sets in $\Bbb{R}^2$? Jun17 asked Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$ Jun17 comment Topologist's sine curve is connected Doesn't disconnected mean that there are two disjoint open sets $A$ and $B$ such that the entire set $S$ is equal to $A\cup B$? But when you take your definitions of $A$ and $B$, the union of them is more than what is needed, no?