If K š†R is Compact and F K 7 †R is Continuous Then F K is Compact
Get the answer to your homework problem.
Try Numerade free for 7 days
Problem Let f: R = R. Show that if f is continuous, then { € R: f(r) < 0} is open: Complete the following proof showing if f is continuous, then for all A € R we have f() € f(A) by filling in the blanks: Let e T then either € A 0r € € A' |^ If € € 4, then f(r) € f(A) € f(). If € € A' | A then pick sequence (Tn)nz1 in with each Tn 7 I and Tn Since f is continuous we have f (Tn) If f(c) € f(A); then f(r) € f(A) € f(A). If f(r) 4 f(A); then each f(Tn) # then the point f(z) is a of f(A): Hence f(z) € f(A).
Related Question
Let U⊂Rn be open and f:U→Rm be a continuous function on U. Prove or disprove each of the following statements: (a) If O⊂U is an open set in Rn, then f(O) is open in Rm. (b) If B⊂U is a bounded set in Rn, then f(B) is bounded in Rm. (c) If {xk} is a Cauchy sequence in U, then {f(xk)}is a Cauchy sequence in Rm.
Discussion
You must be signed in to discuss.
Video Transcript
in this problem given that you said is a subset of R and mm hmm open and function you implies R M B r continuous. This is given and it is continuous on you. So in a part we have to tell that the statement is prove or not or disapprove. Okay, so or a subset of you B open said we have to prove this. So function of you implies are empty. B defined as if X is equals to see and X it implies. So where C is the subset of R M is fixed number, real numbers. Okay, so since our M we have R M is an Euclidean domain. So the single tone set of C cannot be opening because it is single tone set. Okay. Therefore o is not upset All subsets you is an open set in RM then F0 is not opening our M. So this is disapproved. Now we have be part where FB function B is on the bounded offer sequence XN the stock sequence in B. Such that function of X and goes to infinity. That means a sequence affection is divergent because it goes to infinity. Okay, so since b is founded in the tour um we have F X NK is a convergent sequence. Convergent sequence means if talk sequence approaches some limit. Okay then it is called convergent sequence. Okay, so it is in contradiction because there are no limits. Okay, therefore we conclude that be the subset of you is bounded set in at and then I f b is a bounded in our um so this is true. Mrs Okay. Now see part We have led zero is not a subset of you subset of art and be open said. So function you implies to RM is defined by FX is equal to one by X. So let's XN is equals to one by and clearly accent is because she sequence for she sequence means in the sequence the elements are close to each other. Okay? So FXN is equal to one by one by N because N is equals to one by end. So is equal strength, which is not cautious. So this is not Okay, So that's our answer. Thank you.
Source: https://www.numerade.com/ask/question/problem-let-f-r-r-show-that-if-f-is-continuous-then-r-fr-0-is-open-complete-the-following-proof-showing-if-f-is-continuous-then-for-all-a-r-we-have-f-fa-by-filling-in-the-blanks-let-e-t-then-65818/
0 Response to "If K š†R is Compact and F K 7 †R is Continuous Then F K is Compact"
إرسال تعليق