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Does $f$ continuous in the Zariski topology imply that $\underset{x \to x_0}{\lim}f(x)=f(x_0)$?

I used the above argument in a proof, but I found it really suspect and questionable and would like to know whether it is correct or if (and how) to replace it with a correct argument.

If it is correct, would it be possible to sketch a proof, or to point to a reference giving the proof?

Context: Given a projective variety $V$ in $\mathbb{CP}^n$, and two rational functions $\frac{f_1}{g_1}$ and $\frac{f_2}{g_2}$ which are equal everywhere in their common domain of definition, i.e. $\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)$.

I want to show that this implies that $f_1g_2 - f_2g_1$ equals $0$ on every point of $V$.

Obviously this holds everywhere on $\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)$. Then my suspicious argument goes that: this set is Zariski open, hence Zariski dense, in $V$, and $f_1g_2 - f_2g_1$ is a Zariski-continuous function which equals $0$ everywhere in this Zariski dense subset. Thus, because of continuity (the claim in yellow above), it also equals zero on $V \setminus \left[\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)\right]$, thus on all of $V$.

I wanted to argue this more directly, but this was the only idea I could think of at the time.

Note: this corresponds to one direction of exercise 5.4.4. in this book. Since this problem is so basic, I am quite concerned about my current lack of understanding.

Chill2Macht
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    Continuity always implies the limit property. For every open set $U\ni f(x_0)$ there is an open set $V=f^{-1}(U)\ni x_0$ such that for all $x\in V$ one has $f(x)\in U$. However, limits might not be unique in every topology. So, having a continuous function equal to zero in a dense set might not imply that it is constant. – Bettybel Jul 05 '17 at 16:16
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    For example, take $\chi_{{0}}:\mathbb{R}\to {0,1}$ to be the indicator function of ${0}$, put in $\mathbb{R}$ the usual topology and in ${0,1}$ the indiscrete topology. This function is continuous and it is equal to zero in a dense set of $\mathbb{R}$. But it is non-constant. – Bettybel Jul 05 '17 at 16:20
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    By the way, Zariski topology is just definiology. It is only to write prettier statements, but really adds no substance. Substance is, for example, the identity principle of analytic functions. Can you get analytic functions on $\mathbb{C}^{n+1}$ out of your $f_1g_2-f_2g_1$ on $\mathbb{CP}^n$? – Bettybel Jul 05 '17 at 16:37
  • @Bettybel Oh by using the affine charts? And the argument goes through since Zariski open implies Euclidean open? Does the identity principle still works for functions $\mathbb{C}^{n(+1)} \to \mathbb{C}$? I've only studied single-dimensional complex analysis, functions $\mathbb{C} \to \mathbb{C}$. (As an aside, I think the argument is supposed to work for any algebraically closed field $k$, but I always imagine that $k = \mathbb{C}$ since that is the most commonly used in practice and also the algebraically closed field I am most familiar with.) – Chill2Macht Jul 05 '17 at 16:46
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    It works. Not verbatim, though. In one dimension it is enough to have a sequence of points that accumulates. In more dimensions $p(z_1,z_2)=z_1$ vanishes in the whole line $z_1=0$ and it is non-zero. – Bettybel Jul 05 '17 at 16:52
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    In one dimension the identity principle can be proven by proving that the first coefficient must be zero, and applying induction after dividing by $z$. In higher dimension division would need to be replaced by Weierstrass preparation theorem. – Bettybel Jul 05 '17 at 16:55

1 Answers1

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As per the remarks, it is not always true that one may find unique limits, depending on the topology one chooses. I'd like to present a reworking of your argument:

The locus where $f_1g_2-f_2g_1$ vanishes determines a closed subset of $\Bbb P^n$. Since $f_1g_2-f_2g_1$ is zero on $U:=\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)$, the function $f_1g_2-f_2g_1$ must be zero on the closure of $U$, since the closure of $U$ is the smallest closed set containing $U$. But $\overline{U}=V$, and so $f_1g_2-f_2g_1$ is zero on the whole of $V$.

This belies a fun exercise you will probably meet later in life once you know of the word "scheme" and a few adjectives about schemes: any two maps from a reduced scheme to a separated scheme which agree on a dense open are actually equal. The technique of proof is similar: show some locus is closed and contains a dense open.

KReiser
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  • Can I check to see if I understood the argument? $f_1g_2 - f_2 g_1$ vanishes on a closed set $C$ (in topology of $\mathbb{P}^2$), it must also vanish on all subsets of this closed set, which must include the closure of $U$, because $U$ is a subset of $C$, and $\bar{U}$ is the smallest closed subset (in the topology of $\mathbb{P}^2$ or of $V$? or does it not matter? -- it doesn't matter because $V$ is closed in the topology of $\mathbb{P}^2$), so if $\bar{U} \not \subseteq C$, then $C \cap \bar{U}$ would be a smaller closed set containing $U$, which would be a contradiction. Is that correct? – Chill2Macht Jul 06 '17 at 13:16
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    Yes, that is correct. – KReiser Jul 06 '17 at 17:29