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I encountered some difficulties with Problem 13 from Chapter 6 of Spivak's calculus dealing with continuities. The question in whole:

(a) Prove that if $f$ is continuous on $[a, b]$, then there is a function g which is continuous on $\mathbb{R}$, and which satisfies $g(x) = f(x)$ for all $x$ in $[a, b]$. Hint: Since you obviously have a great deal of choice, try making g constant on $(-\infty, a]$ and $[b, \infty)$.

(b) Give an example to show that this assertion is false if $[a, b]$ is replaced by $(a,b)$.

I understand the solution to part (a), which gives the following function for $g(x)$:

$$ g(x) = \left\{ \begin{array}{ll} \lim_{x \to a^+} f(x) & \mbox{if } x \le a \\ f(x) & \mbox{if } a < x < b \\ \lim_{x \to b^-} f(x) & \mbox{if } x \ge b \end{array} \right. $$

My interpretation was that since $f$ is continuous on $[a,b]$, then,

$$ \lim_{x \to a^+} f(x) = f(a) \\ \lim_{x \to b^-} f(x) = f(b) $$

Hence, the function definition for $g(x)$ satisfies $g(x) = f(x)$ for all $x$ in $[a, b]$.

However, for part (b), I just don't understand how the assertion becomes false just by replacing the closed interval $[a, b]$ with the open interval $(a, b)$. The only change I can visualize is that we can no longer make the initial statement that $\lim_{x \to a^+} f(x) = f(a)$ and $\lim_{x \to b^-} f(x) = f(b)$. However, I don't understand how is it then that $g(x) = f(x)$ cannot still hold for $a < x < b$.

Any pointers would be greatly appreciated!

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  • $\begingroup$ You're on the right track: the issue is that $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ may not necessarily exist anymore (so your initial proposal for $g$ is not even well-defined anymore). So, to come up with an explicit counterexample, can you think of a function for which the limit at a certain point does not exist? (Hint: read through the text again, you'll definitely find an example) $\endgroup$
    – peek-a-boo
    Aug 25, 2020 at 1:12
  • $\begingroup$ What makes the assertion false is not that $g(x) = f(x)$ can't hold for $x\in(a,b)$ but that, unlike the first case, $g$ no longer exists for all possible $f$. Take $f(x) = \frac{1}{x-a}$ and try to prove that for any value we choose for $g(a)$, $g$ won't be continuous at $x=a$. $\endgroup$
    – gabrimev
    Aug 25, 2020 at 1:20
  • $\begingroup$ Oh no, I see. I was under the impression that the function $g(x)$ could be 're-defined' to simply $g(x) = f(x)$ with the change to the interval. I see now that the original $g(x)$ is rendered invalid regardless of the value a. Thank you! $\endgroup$
    – iobtl
    Aug 25, 2020 at 1:34
  • $\begingroup$ @iobtl for what it's worth: I think that part of the difficulty that you had was caused because part(b) of the problem was worded in a valid but tricky manner. I would have worded part (b) : show that is possible to construct a function $f$ that is continuous on $(a,b)$ $\{$that is, $f$ may or may not be continuous on $[a,b]\}$ and such that no satisfying function $g$ can be constructed. $\endgroup$ Aug 25, 2020 at 5:52
  • $\begingroup$ @user2661923 yep, that may have been part of it. All is well though, I understand the key idea the problem was trying to bring across with regards to the continuity in a closed vs open interval. $\endgroup$
    – iobtl
    Aug 25, 2020 at 13:07

4 Answers 4

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As mentioned in the comments, $(b)$ asks to construct a function $f$ that is continuous on $(a,b)$ for which there is no function $g$ satisfying

  1. $g(x)=f(x)$ for all $x$ in $(a,b)$.
  2. $g$ is continuous on $\mathbb R$.

Since the closed interval $[a,b]$ is replaced by the open interval $(a,b)$, the function $f$ doesn't need to be continuous at the endpoints $a$ and $b$.

Two valid constructions for $f$ are:

$$f(x)=\frac{1}{x-a} \quad {\text{or}}\quad f(x)=\frac{1}{x-b}.$$

Both of these satisfy the condition that $f$ is continuous on $(a,b)$. However, if $g(x)=f(x)$ on $(a,b)$ then $g$ becomes arbitrarily large around $a$ or $b$ and therefore cannot be continuous on $\mathbb R$.

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  • $\begingroup$ Extremely clear and straightforward. Thank you! $\endgroup$
    – iobtl
    Aug 26, 2020 at 1:20
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Take $f(x) =\dfrac{1}{x}$ which is continuous on $(0,1)$, can you find such $g(x)$?

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The reason why the extension works in part (a) is because $[a,b]$ is a closed interval, and $f$ is a continuous function, so

1.$\lim_{x\rightarrow a}f(x)$ exists and is finite.

2.$\lim_{x\rightarrow a}f(x)=f(a)$.

However, if we define $\tan(x)$ on $(-\frac{\pi}{2},\frac{\pi}{2})$, then both 1. and 2.above fail, since $$\lim_{x\rightarrow \pi/2}\tan(x)=\infty\text{ and }\lim_{x\rightarrow -\pi/2}\tan(x)=-\infty.$$ So in these sort of cases, no continuous extension can exist. Hopefully this helps.

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A function may be strictly increasing on the interval $(a,b)$ and approach $+\infty$ as $x$ approaches $b$ from below and approach $-\infty$ as $x$ approaches $a$ from above.

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