## anonymous one year ago please proving continuity help

1. imqwerty

???

2. 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]

3. anonymous

prove that F is continues

4. anonymous

@zzr0ck3r

5. anonymous

@Loser66

6. 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 $$x_0\in [2,4]$$

7. 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.

8. anonymous

thank you sir

9. zzr0ck3r

@Loser66 you proved for $$f(x) = x$$ We need to prove if for $f(x) = \begin{cases} 1 & 0\le x\leq 1 \\ 2 & 2\leq x\leq 4 \end{cases}$

10. Loser66

Please, fix it. :) Much appreciate.

11. zzr0ck3r

I need to know what form he wants, I think, because of the subspace, this is a topology question.

12. 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.

13. zzr0ck3r

Make sense?