Limits of functions and sequences

[S.G. Janssens]

Or is ze risky ( Zariski ;) ) topology the same in the spaces in question as the finite complement topology?

On the real line this is true, because there is a one-to-one correspondence between zero sets of polynomials and finite sets. Whether this is true in full generality, I do not know by heart. Probably @mathwonk knows.

 

[WWGD]

On the real line this is true, because there is a one-to-one correspondence between zero sets of polynomials and finite sets. Whether this is true in full generality, I do not know by heart. Probably @mathwonk knows.

Actually, I was just thinking about it and the zero locus over 2+ dimensions can be ( uncountably) infinite, as in $x^2+y^2+z^2-1$ so likely the two topologies are not equal.

 

[S.G. Janssens]

Actually, I was just thinking about it and the zero locus over 2+ dimensions can be ( uncountably) infinite, as in $x^2+y^2+z^2-1$ so likely the two topologies are not equal.

I don't think multivariates qualify, because if they would qualify, then I do not see how every sequence consisting of distinct points would have to converge to every point in the space:

You have made me realize another odd fact about the zariski topology that we use in algebraic geometry. On any affine space, say the real line, zariski closed sets are zero loci of polynomials, hence either the whole line, or finite sets. Hence every sequence consisting of distinct points, converges in this topology to every point of the line. Needless to say we don't use sequences much in this topology.

 

[mathwonk]

Very observant. Indeed what seems to be true in more variables is that any sequence of distinct points that eventually lies entirely on an algebraic curve, converges to every point of that curve.

 

[S.G. Janssens]

Very observant. Indeed what seems to be true in more variables is that any sequence of distinct points that eventually lies entirely on an algebraic curve, converges to every point of that curve.

Suppose that $C$ and $D$ are two distinct algebraic curves in the (real) plane and let $x \in C \setminus D$. If I understand you correctly, then any sequence of distinct points in $C$ should eventually be in $C \setminus D$, for that sequence to converge to $x$. However, what if $C$ and $D$ have a common component $E$ and the sequence in question lies entirely in $E$?

I understand that the case of $C$ and $D$ having a common component is "non-generic", but the definition of convergence requires that it works in this case, too, doesn't it?

 

[mathwonk]

excellent point! you are quite right that one should consider also possibly reducible curves, but i tend to think of irreducible ones. So, good point well taken, a closed subset of a reducible curve may not be finite, it might be a whole component! so maybe add "irreducible" to my statements about curves.

 

[S.G. Janssens]

but i tend to think of irreducible ones.

I think this makes a lot of sense, given your background.

Thank you, and also @WWGD, for the little discussion.