Left continuity

Showcase_22 · #1 $Old$ May 30th, 2009, 02:08 PM

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is increasing and satisfies the conclusion of the intermediate value theorem, then prove f is left continuous.

My attempt:

$\forall r \in [f(a),f(b)] \exists c \in [a,b] \ s.t \ f(c)=r.$

Hence pick $b=a+\epsilon$ and let $\epsilon=\epsilon_n$ s.t $\epsilon_n=1/n$ .

Let $r_n=f(a+\epsilon_n)$ .

Also let $f(a)=c_1$ (just to make it different from the definition).

therefore $\lim_{n \rightarrow \infty} f(a+\epsilon_n)\rightarrow f(r_{\infty})=f(a)=c_1$

Hence it is left continuous.

Is this right? I have a feeling that i've assumed it to prove it

Sampras · #2 $Old$ May 30th, 2009, 02:19 PM

Quote:

Originally Posted by Showcase_22 $View Post$

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is increasing and satisfies the conclusion of the intermediate value theorem, then prove f is left continuous.

My attempt:

$\forall r \in [f(a),f(b)] \exists c \in [a,b] \ s.t \ f(c)=r.$

Hence pick $b=a+\epsilon$ and let $\epsilon=\epsilon_n$ s.t $\epsilon_n=1/n$ .

Let $r_n=f(a+\epsilon_n)$ .

Also let $f(a)=c_1$ (just to make it different from the definition).

therefore $\lim_{n \rightarrow \infty} f(a+\epsilon_n)\rightarrow f(r_{\infty})=f(a)=c_1$

Hence it is left continuous.

Is this right? I have a feeling that i've assumed it to prove it

It seems that you should consider $r_n = f(a-\epsilon_n)$ since you want to show left continuity.

Showcase_22 · #3 $Old$ May 30th, 2009, 02:24 PM

$\forall r \in [f(a),f(b)] \exists c \in [a,b] \ s.t \ f(c)=r.$

Hence pick $b=a-\epsilon$ and let $\epsilon=\epsilon_n=1/n$ .

Let $r_n=f(a+\epsilon_n)$ .

Since f is an increasing sequence $r_1>r_2>r_3......$ .

f is bounded below by $f(a)$ . Therefore $r_{\infty}=f(a)$ .

Also let $f(a)=c_1$ (just to make it different from the definition).

therefore $\lim_{n \rightarrow \infty} f(a-\epsilon_n)\rightarrow f(r_{\infty})=f(a)=c_1$

Hence it is left continuous.

Is that better?

Sampras · #4 $Old$ May 30th, 2009, 02:36 PM

Quote:

Originally Posted by Showcase_22 $View Post$

$\forall r \in [f(a),f(b)] \exists c \in [a,b] \ s.t \ f(c)=r.$

Hence pick $b=a-\epsilon$ and let $\epsilon=\epsilon_n=1/n$ .

Let $r_n=f(a+\epsilon_n)$ .

Since f is an increasing sequence $r_1>r_2>r_3......$ .

f is bounded below by $f(a)$ . Therefore $r_{\infty}=f(a)$ .

Also let $f(a)=c_1$ (just to make it different from the definition).

therefore $\lim_{n \rightarrow \infty} f(a-\epsilon_n)\rightarrow f(r_{\infty})=f(a)=c_1$

Hence it is left continuous.

Is that better?

I would let $r_n = f(a-\epsilon_n)$ . Then $r_n$ is an increasing sequence. And so $r_{\infty} = f(a)$ . Now taking $r = a, \ c = c_1$ in the Intermediate Value Theorem, we get $f(a) = c_1$ . So $\lim_{n \to \infty} f(a-\epsilon_n) = r_{\infty} = f(a) = c_1$ .

xalk · #5 $Old$ June 3rd, 2009, 03:41 AM

Quote:

Originally Posted by Showcase_22 $View Post$

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is increasing and satisfies the conclusion of the intermediate value theorem, then prove f is left continuous.

My attempt:

$\forall r \in [f(a),f(b)] \exists c \in [a,b] \ s.t \ f(c)=r.$

Hence pick $b=a+\epsilon$ and let $\epsilon=\epsilon_n$ s.t $\epsilon_n=1/n$ .

Let $r_n=f(a+\epsilon_n)$ .

Also let $f(a)=c_1$ (just to make it different from the definition).

therefore $\lim_{n \rightarrow \infty} f(a+\epsilon_n)\rightarrow f(r_{\infty})=f(a)=c_1$

Hence it is left continuous.

Is this right? I have a feeling that i've assumed it to prove it

Showcase, the intermediate value theorem says:

If f(x) is continuous in [a,b] and if f(a) =A and f(b)=B,then corresponding to any number C between A and B there exists at least one number c in [a,b] such that f(c) =C.

So it seems to me you are trying to prove the basic assumption of the intermediate value theorem,since if a function is continuous then is left,right continuous