Timeline for Extending a uniformly continous function to the closure of its domain
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:20 | history | edited | CommunityBot | replaced http://math.stackexchange.com/ with https://math.stackexchange.com/ | |
| Dec 17, 2014 at 21:04 | comment | added | Ayman Hourieh | @Mike I didn't spell out all of the details. This is indeed what you need to do. To find $\hat x$, you have two deltas, you take the minimum and then pick $\hat x$ within the this minimum. The same applies to $\hat y$. | |
| Dec 17, 2014 at 20:29 | comment | added | Mike | Sorry for the bother, but I think something in the proof is wrong. Continuity means that given an epsilon you can find delta; You have said that the specific delta satisfies the condition. Shouldn't it be another delta, and then take the minimum of them? | |
| Dec 15, 2014 at 18:59 | vote | accept | Mike | ||
| Dec 15, 2014 at 18:58 | comment | added | Ayman Hourieh | @Mike As to your question, that follows from the continuity of $f$ at $x$. I'm not using uniform continuity here. And yes, $\hat x$ is fixed for this $x$. | |
| Dec 15, 2014 at 18:52 | comment | added | Mike | Thanks a lot!!! Question - why can you conclude from $d(x, \hat x) < \delta / 3$ that $d(f(x), f(\hat x)) < \epsilon/3$? Isn't this just using excatly what we need to prove? or is $\hat x$ being constant? | |
| Dec 15, 2014 at 18:41 | history | answered | Ayman Hourieh | CC BY-SA 3.0 |