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Mar 7, 2016 at 18:23 vote accept Jo Wehler
Mar 7, 2016 at 18:20 comment added Jo Wehler @Todd Trimble I made an edit: The closure of the support is a compact subset of U.
Mar 7, 2016 at 18:18 history edited Jo Wehler CC BY-SA 3.0
added: relative-compact
Mar 7, 2016 at 18:12 comment added Todd Trimble I had trouble understanding this at first. But I guess $\mathcal{F}(U)$ is by definition the set of continuous functions whose support is contained in $U$?
Mar 7, 2016 at 18:05 answer added Simon Henry timeline score: 9
Mar 7, 2016 at 17:43 review First posts
Mar 7, 2016 at 18:01
Mar 7, 2016 at 17:42 history asked Jo Wehler CC BY-SA 3.0