If $f$ is continuous on $\mathbb R$, is $f$ also continuous on $\mathbb R-\{0\}$?

Question

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.)

Answer 1

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 $\mathbb{R}$.

Answer 2

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 $\Bbb R$.

Answer 3

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).