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May 4, 2014 at 13:44 history edited Partha Pratim Ghosh CC BY-SA 3.0
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Apr 30, 2014 at 17:44 comment added Partha Pratim Ghosh Also, I could not follow the proof of your statement; could you please provide a proof?
Apr 30, 2014 at 17:43 comment added Partha Pratim Ghosh Thanks Ramiro, however, if one takes the identity map on $I(B)$, then also $P$ and $G$ are homeomorphic; in particular, if $B$ were zero dimensional then its identity map is such a case. But then $f$ is not constant, and neither does every open subset become closed.
Apr 30, 2014 at 16:53 comment added Ramiro de la Vega If $f$ is constant then both $P$ and $G$ are homeomorphic to $X$. Thus $I(P)$ and $G$ are homeomorphic only if $X$ is such that every open set is closed. In the $T_0$ case this means that $X$ is discrete.
Apr 30, 2014 at 14:21 review First posts
Apr 30, 2014 at 14:46
Apr 30, 2014 at 14:04 history asked Partha Pratim Ghosh CC BY-SA 3.0