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Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x) = x^2$. I have a rather elementary question, but can one say that this function is continuous on the whole real line?

If we restrict the codomain to $\mathbb{R}_{\geq 0}$ then continuity is clear using both the $\epsilon$-$\delta$ definition or the topological one. However for negative image points this may be a problem. I believe we may circumvent this by saying $f^{-1}((-\infty, 0)) = \emptyset$, which is open by definition, and hence the map is overall continuous. However it seems silly to say a function is continuous at points/sets where it cannot be defined. Moreover, trick will not work using the $\epsilon$-$\delta$ definition and so it appears to be incorrect.

CBBAM
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  • @ThomasAndrews Yes that was a typo, my apologies. – CBBAM Aug 31 '22 at 18:26
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    “However it seems silly to say a function is continuous at points/sets where it cannot be defined” – but $f(x) = x^2$ is defined everywhere. – Martin R Aug 31 '22 at 18:28
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    “Moreover, trick will not work using the $\epsilon$-$\delta$ definition and so it appears to be incorrect.” – unclear to me as well. The $\epsilon$-$\delta$ definition for continuity works without problems, at any $x \in \Bbb R$. – Martin R Aug 31 '22 at 18:29
  • First of all, "continuous on the whole real line" is about the domain, while $(-\infty,0)$ is an open subset of the range. – Thomas Andrews Aug 31 '22 at 18:32

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First, it's not a trick per se to write $f^{-1}((-\infty,0)) = \emptyset$, there are no values $x \in \mathbb{R}$ such that $f(x) = x^2 \in (-\infty,0)$, so the preimage is empty. You are correct then in saying that using the topological definition of continuity the function $f$ is continuous on $\mathbb{R}$ with the standard topology.

The function is also continuous using the the $\epsilon-\delta$ definition. Remember that the $\epsilon-\delta$ definition of continuity says that for any $x_0 \in Dom(f)$ and any $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x-x_0| < \delta$ then $|f(x)-f(x_0)|<\epsilon$. Since there are no $x_0 \in \mathbb{R}$ such that $f(x_0) < 0$, there is no problem with negative values in the codomain.

perpetuallyconfused
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  • Ah I see now, I believe it was a silly mistake on my end to not write out the $\epsilon$-$delta$ definition for this problem. Thanks! – CBBAM Aug 31 '22 at 18:33
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We say a function is continuous at points in its domain, not its codomain. In noting that $f^{-1}((-\infty, 0)) = \emptyset$, you have show that $f$ is continuous on $\emptyset$ which is, since $\emptyset$ is open, vacuously true.

DanishChef
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Re the general definition, that $f$ is continuous iff ( def.) the inverse image of an open set is open, consider any open $(a,b)$ set in the "Target" $\mathbb R_+$ (i.e., so that $0\leq a< b $). Its inverse image under $f$ is $( \sqrt a, \sqrt b) \cup (-\sqrt a, -\sqrt b)$, which is the union of two ope sets, ad is therefore open in $\mathbb R$

MSIS
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