The Klub 17 Sequence4/10/2021
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The Klub 17 Sequence How To Fucking EatI know you learned to eat a fucking brunch, but if you havent, I will show you how to fucking eat it RIP pic.twitter.comhIdxfq9bqG.Vintage Porn Collector Links ( ) Vintage Porn Collector Links - xNovel The following sites may be of interest to collectors of vintage porn. What we have got is a sequence xnk to x in I and f(xnk) to infty; (because f(xnk) gt nk). ![]() For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval x0,x0P. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above real-analysis continuity uniform-continuity periodic-functions share cite improve this question edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 endgroup add a comment 17 6 begingroup A function f:mathbbRto mathbbR is periodic if there exits p0 such that f(xP)f(x) for all xin mathbbR. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above real-analysis continuity uniform-continuity periodic-functions share cite improve this question edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 endgroup add a comment 17 6 17 6 17 6 begingroup A function f:mathbbRto mathbbR is periodic if there exits p0 such that f(xP)f(x) for all xin mathbbR. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above real-analysis continuity uniform-continuity periodic-functions share cite improve this question edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 endgroup A function f:mathbbRto mathbbR is periodic if there exits p0 such that f(xP)f(x) for all xin mathbbR. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above real-analysis continuity uniform-continuity periodic-functions real-analysis continuity uniform-continuity periodic-functions share cite improve this question edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 share cite improve this question edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 share cite improve this question share cite improve this question edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 edited Jan 14 at 13:07 Martin Sleziak 44.8k 9 118 272 44.8k 9 118 272 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 asked Apr 30 14 at 1:13 Oscar Flores Oscar Flores 623 6 18 623 6 18 add a comment add a comment 3 Answers 3 active oldest votes 16 begingroup Suppose that f has period one. Since f:0,2toBbb R is continuous, it is bounded, so f is bounded all over Bbb R (why). Also, f:0,2toBbb R is uniformly continuous, being continuous on a compact set. Thus, given varepsilon 0 there exists delta0 such that, whenever x-y Drawing a picture would prove useful. Essentially, youre translating the problem to 0,2 where we already solved the issue. A continuous function on a compact space is both bounded and uniformly continuous. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle. Since xn is bounded and consists of real numbers, it has a convergent subsequence xnk. But I is closed hence xnk converges inside I. I such that xnk to x.
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