please proving continuity help

Question

please proving continuity  help

imqwerty · Answer

???

anonymous · Answer

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 · Answer

prove that F is continues

anonymous · Answer

@zzr0ck3r

anonymous · Answer

@Loser66

loser66 · Answer

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 · Answer

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 · Answer

thank you sir

zzr0ck3r · Answer

@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 · Answer

Please, fix it. :) Much appreciate.

zzr0ck3r · Answer

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

zzr0ck3r · Answer

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 · Answer

Make sense?