Timeline for Why don't clopen sets pose problems in the "preimage" definition of continuity, and in the definition of a topology?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2020 at 10:38 | history | edited | CommunityBot | Commonmark migration | |
| S Jan 4, 2016 at 16:56 | history | edited | Em. | CC BY-SA 3.0 | new title; some formatting changes; removed unnecessary "sorry if this is a bad question" opening |
| S Jan 4, 2016 at 16:56 | history | suggested | i'm nobody who are you | CC BY-SA 3.0 | new title; some formatting changes; removed unnecessary "sorry if this is a bad question" opening |
| Jan 4, 2016 at 16:38 | review | Suggested edits | |||
| S Jan 4, 2016 at 16:56 | |||||
| Jan 2, 2016 at 4:24 | vote | accept | AnotherPerson | ||
| Jan 2, 2016 at 4:23 | comment | added | user160738 | What's important about continuity definition of function in terms of preimage is that it is equivalent to classic definition of $\epsilon$-$\delta$ continuity. Another equivalent definition is: a function is continuous if every preimage of open sets is also open. These definitions are useful because it only makes use of topological concepts, so can be used for general functions where notion of distance may not necessarily be defined. | |
| Jan 2, 2016 at 4:16 | answer | added | David Wheeler | timeline score: 1 | |
| Jan 2, 2016 at 4:16 | answer | added | Alex Provost | timeline score: 2 | |
| Jan 2, 2016 at 4:09 | history | asked | AnotherPerson | CC BY-SA 3.0 |