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I mean, if I make the latter claim, am I precluding the possibility that $f$ could actually be continuous on the entire reals? (I am right now proving the continuity of a 2-piece function joined in a continuous fashion at 0, and I've just made the latter claim before I go on to show that f is continuous at 0 and hence over the entire reals. But on re-reading, my claim seems to misleadingly suggest that 0 is a discontinuity.)

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I think that if a function is continuous over a set, then it is also continuous over a contained set –  Alan Simonin Jan 19 '13 at 16:59
    
A function is continuous on a set if it is continuous at every point of that set. –  Sigur Jan 19 '13 at 17:02
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The answer is yes. Note that often in proofs we prove that $f$ is continuous on a set by proving that it is continuous on some subsets with the union the entire set. –  N. S. Jan 19 '13 at 17:03
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You might tend to assume that $0$ is a point of discontinuity, but you would do so knowing that it might not be. –  Neal Jan 19 '13 at 17:04
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@Neal Haha, thanks, we are at exactly at the same frequency then! –  Ryan Jan 19 '13 at 17:08

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If $f$ is continuous on $\mathbb{R}$, then $f$ is continuous at each $x\in\mathbb{R}$. Since each $x\in\mathbb{R}-\{0\}$ is also an element of $\mathbb{R}$, then $f$ is certainly continuous on $\mathbb{R}-\{0\}$.

If you make the latter claim, you do not preclude the possibility that $f$ is continuous on $\mathbb{R}$. It boils down to the difference between "does not need to be" and "needs to not be". If $f$ is continuous everywhere but $0$, then $f$ does not need to be continuous on all of $\mathbb{R}$; that doesn't mean it must not be continuous on all of $\mathbb{R}$.

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Thanks for the reassurance. –  Ryan Jan 19 '13 at 17:15

You are correct. If $f$ is continuous on $\Bbb R\smallsetminus\{0\}$, then it need not be continuous on $\Bbb R$--consider $$f(x)=\begin{cases}1 & x>0\\ 0 & x=0\\-1 & x<0,\end{cases}$$ for example--however, if you are also given that $f$ is continuous at $0$, then it does follow that $f$ is continuous on $\Bbb R$.

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You are right; a function continuous on $\Bbb R-\{0\}$ need not be continuous on $\Bbb R$, as $$f(x)=\begin{cases}\frac{1}{x} &:\ x\neq0\\ \\ 0 &:\ x=0\end{cases}$$ shows. However, if you know $$\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=L,$$ then you can extend continuity to $\Bbb R$ by defining $f(0):= L$ or verifying that the limits match the value of the function (such as in verifying piecewise-continuous functions).

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