Timeline for Non trivial definition of bicontinuous functions and the ring of all bicontinuous functions.
Current License: CC BY-SA 3.0
9 events
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| Sep 9, 2012 at 17:31 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Sep 9, 2012 at 4:49 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Aug 4, 2012 at 6:09 | comment | added | Ali Reza | Consider two topologies $\tau_1$={$(a,\infty):a\in \mathbb{R}$} and $\tau_2$={$(-\infty,b):b\in \mathbb{R}$} on the real numbers. the identity function is a continuous function from $(\mathbb{R},\tau_1 ,\tau_2)\rightarrow (\mathbb{R},\tau_1 ,\tau_2)$. But its obvious that −f is not continuous. this implies that the set of all such bicontinuous functions with your definition does not form a ring. How are you sure that your definition implies that it forms a ring | |
| Aug 4, 2012 at 4:12 | comment | added | David Feldman | What happens if you consider two different topologies on $\Bbb R$, say $t_1$ and $t_2$ and consider functions that are both $\tau_1$ to $t_1$ continuous and $\tau_2$ to $t_2$ continuous? Such functions obviously form a ring, but I don't see yet how to choose everything to get an interesting example. | |
| Aug 3, 2012 at 21:07 | comment | added | Ali Reza | We Know that if we define bicontinuous function $f:(X,\tau_1 ,\tau_2) \rightarrow \mathbb{R}$ so that $ f:(X,\tau_1) \rightarrow \mathbb{R}$ and $f:(X ,\tau_2) \rightarrow \mathbb{R}$ are continuous, then we have a pair of rings. i.e.$C(X,\tau_1)$ and $C(X,\tau_2)$. which it seems that there is no general relation between these rings which followed by bicontinuous definition of it. But it is trivial to consider the situations with which the set of all bicontinuous functions forms a ring. | |
| Aug 3, 2012 at 20:48 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Aug 3, 2012 at 20:46 | comment | added | Qiaochu Yuan | Why a ring and not a pair of rings? | |
| Aug 3, 2012 at 19:40 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Aug 3, 2012 at 19:29 | history | asked | Ali Reza | CC BY-SA 3.0 |