anonymous
  • anonymous
please proving continuity help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
imqwerty
  • imqwerty
???
anonymous
  • anonymous
let x and y be subspace of R given by x=[0,1]U[2,4] defined by f: x--->R by f(x)=1 : x E [0,1] and 2 :x E [2,4]
anonymous
  • anonymous
prove that F is continues

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
@Loser66
Loser66
  • Loser66
let fix \(x_0\in [0,1]\), By definition, we must find \(\delta >0\) for given \(\epsilon >0\) such that |\(||x-x_0||<0\) implies \(||f(x)-f(x_0)||<\epsilon \). Choose \(\delta=\epsilon\), the definition becomes \(||x-x_0||<\delta\) implies \(||x-x_0||<\epsilon\). Hence f is continuous. Do the same with \(x_0\in [2,4]\)
Loser66
  • Loser66
Other proof is valid also, since [0,1] and [2,4] are compact sets, hence f is continuous but I am not sure whether you study compact sets or not.
anonymous
  • anonymous
thank you sir
zzr0ck3r
  • zzr0ck3r
@Loser66 you proved for \(f(x) = x\) We need to prove if for \[f(x) = \begin{cases} 1 & 0\le x\leq 1 \\ 2 & 2\leq x\leq 4 \end{cases}\]
Loser66
  • Loser66
Please, fix it. :) Much appreciate.
zzr0ck3r
  • zzr0ck3r
I need to know what form he wants, I think, because of the subspace, this is a topology question.
zzr0ck3r
  • zzr0ck3r
Suppose \(U\) is opoen in \(\mathbb{R}\). Then \(U= \cup_{i\in I}B_i\) for some index \(I\) where the \(B_i\) are intervals of the form \((a,b)\). Now \(f^{-1}(\cup_{i\in I}B_i)=U_{i\in I}f^{-1}(B_i)\). Consider \(i\in I\). If \(1\in B_i\) and \(2\notin B_i\) then \(f^{-1}(B_i)=[0,1]\) which is open in the subspace \([0,1]\cup[2,4]\) because \([0,1]=(-1, 1.5)\cap [0,1]\cup [2,4]\) If \(2\in B_i\) and \(1\notin B_i\) then \(f^{-1}(B_i)=[2,4]\) which is open in the subspace \([0,1]\cup[2,4]\) because \([2,4]=(1.5, 5)\cap [0,1]\cup [2,4]\). If both \(1, 2\in f^{-1}(B_i)\) then \(f^{-1}(B_i)=[0,1]\cup[2,4]\) which is open in \([0,1]\cup[2,4]\) because it is the entire subspace. Finally, if \(1,2\notin f^{-1}(B_i)\), then \(F^{-1}(B_i)=\emptyset\) which is of course open. So for each \(i\in I\) we have \(f^{-1}(B_i)\) is open and thus so is \(\cup_{i\in I}f^{-1}(B_i)=f^{-1}(\cup_{i\in I}B_i)=f^{-1}(U)\) showing that \(f\) is indeed continuous.
zzr0ck3r
  • zzr0ck3r
Make sense?

Looking for something else?

Not the answer you are looking for? Search for more explanations.