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Let $F: D\rightarrow\mathbb{R}$ and a point $x_0\in\mathbb{R}$. Define $A=\{x\in D| x\geq x_0\}$ and $B=\{x\in D| x\leq x_0\}$. Prove that $F: D\rightarrow\mathbb{R}$ is continuous at $x_0$ iff $f:A\rightarrow\mathbb{R}$ and $F: B\rightarrow\mathbb{R}$ is continuous at $x_0$. I need help with the backwards direction.

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Suppose that $f:A\rightarrow\mathbb{R}$ and $F: B\rightarrow\mathbb{R}$ are continuous at $x_0$. We see that $A\cup B=D$. Let $x_n$ be a sequence in $A\cup B$ such that $x_n\rightarrow x_0$. Then it follows from assumption that $f(x_n)\rightarrow f(x_0)$. Hence $f:D\rightarrow\mathbb{R}$ is continuous at $x_0$. Is this all that I need?

New Proof: Suppose that $f: A\rightarrow\mathbb{R}$ is continous at $x_0$. It follows that if $\epsilon>0$, then there exists a $\delta_1>0$ where $\delta_1=\epsilon$ and $|x-x_0|<\delta_1$ then $|f(x)-f(x_0)|<\epsilon$. Suppose now that $f: B\rightarrow\mathbb{R}$ is also continuous at $x_0$. It follows that if $\epsilon>0$, then there exists a $\delta_2=\epsilon$ and $|x-x_0|<\delta_2$ then $|f(x)-f(x_0)|<\epsilon$. Thus if we let $\delta=min(\delta_1,\delta_2)$ and $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$.

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  • $\begingroup$ Is $D$ a metric space? $\endgroup$ – Stefan Hamcke Dec 25 '13 at 18:22
  • $\begingroup$ No its the domain. $\endgroup$ – user60887 Dec 25 '13 at 18:22
  • $\begingroup$ @StefanHamcke As written, $D \subset \mathbb{R}$. $\endgroup$ – Daniel Fischer Dec 25 '13 at 18:23
  • $\begingroup$ You are just rephrasing the fact that $\lim_{x \to x_0} f(x)=\ell$ if and only if $\lim_{x \to x_0+} f(x) = \ell = \lim_{x \to x_0-}f(x)$. $\endgroup$ – Siminore Dec 25 '13 at 18:24
  • $\begingroup$ So would I need to create two sequences then one in $A$ and one $B$. And show they both converge to $x_0$? $\endgroup$ – user60887 Dec 25 '13 at 18:26
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Notice that, $f:A\rightarrow\mathbb{R}$ is continuous at $x_0$, that means it is continuous from the right, that is

$$ \lim_{x\to x_0^+} f(x)=f(x_0), $$

and similarly, $f:B\rightarrow\mathbb{R}$ is continuous at $x_0$, that means it is continuous from the left

$$ \lim_{x\to x_0^-} f(x)=f(x_0). $$

Can you conclude?

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  • $\begingroup$ Oh then the limit from the right and limit from the left must be equal. By the uniqueness of limits the limit $f(x)=f(x_0)$ exists as $x\rightarrow x_0$. $\endgroup$ – user60887 Dec 25 '13 at 20:07
  • $\begingroup$ @user60887: You have $ \lim_{x\to x_0^+} f(x)= \lim_{x\to x_0^-} f(x)=f(x_0)$ which implies $ \lim_{x\to x_0} f(x)=f(x_0) $. $\endgroup$ – Mhenni Benghorbal Dec 25 '13 at 20:21
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In your first attempt you are only reformulating the claim and throwing in a “we see”. You could just as well “see” that the original claim is true.

Your second attempt is good. I too find the $\epsilon$-$\delta$-criterion easier to work with then sequences for this problem. A minor problem is that you should fix an $\epsilon$ first, and then consider both of the restrictions of $f$, to make sure that it is understood that you are talking about the same $\epsilon$ everywhere. The not so minor problem is that beginning with $\delta_1=\epsilon$ your sentence goes astray. What exactly are you trying to say?

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