Timeline for Topological conditions forcing continuity
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 31, 2014 at 13:42 | vote | accept | Jacques Carette | ||
| Aug 29, 2014 at 11:47 | answer | added | Dominic van der Zypen | timeline score: 11 | |
| Mar 7, 2014 at 0:33 | history | edited | user9072 |
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| Jan 4, 2012 at 20:12 | comment | added | Jacques Carette | Thanks for the expansion. The situation where this arises concerns 'generalized Stream' spaces, $Y = \mathbb{T}\rightarrow Q$ where $\mathbb{T}$ is time-like (might be $\mathbb{R}$, or $\mathbb{R}^+$ but also much weirder spaces), and $Q$ is some kind of (topological) algebra. | |
| Jan 4, 2012 at 18:44 | comment | added | Bill Johnson | The second condition is sufficient with no assumptions on the topologies, Jacques. The first condition is just the observation that the second condition applies when $Y$ is completely regular and $Z=[0,1]$, which I (perhaps wrongly) assumed would take care of any situation that arose in Computer Science. $$ $$ I haven't thought about it, but would guess that the second condition is necessary in order for what you want to be true for all $X$ and all $f$. | |
| Jan 3, 2012 at 20:15 | comment | added | Jacques Carette | @Bill: thanks. But somehow that still seems like asking 'too much' for what you get. For example, why 'regular' but not Hausdorff or something weaker still? | |
| Jan 3, 2012 at 19:32 | comment | added | Bill Johnson | It is enough that $Y$ be homeomorphic to a subset of some product of $Z$, so a simple sufficient condition is that $Y$ be completely regular and $Z=[0,1]$. More generally, it is sufficient that there be enough continuous functions from $Y$ to $Z$ to separate points in $Y$ from disjoint closed subsets of $Y$. | |
| Jan 3, 2012 at 18:30 | history | asked | Jacques Carette | CC BY-SA 3.0 |