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anuttarasammyak

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For any real number x, f(x) is 0 or 1 but lim ##x\rightarrow a## f(x) does not exist for any real number a.

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anuttarasammyak

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anuttarasammyak

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If there is a discontinuity, no limit exists on that point.

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Mark44

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- #8

Stephen Tashi

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If there is a discontinuity, no limit exists on that point.

That's true of the specific function in post #2. However, in general, if ##f(x)## is discontinuous at ##x = a## then it is possible for ##\lim_{x \rightarrow a} f(x) = L ## to exist. The fact ##f(x)## is discontinuous at ##x = a## only says that ##L \ne f(a)##.

- #9

anuttarasammyak

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For an example

f(x)=1 for x ##\neq## 0, 0 for x=0

f(0)=0

lim ##x \rightarrow 0## f(x)=1

f(x)=1 for x ##\neq## 0, 0 for x=0

f(0)=0

lim ##x \rightarrow 0## f(x)=1

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- #11

anuttarasammyak

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[tex] \lim_{x\rightarrow a_i-0}f(x)=b_{i-}[/tex]

[tex] \lim_{x\rightarrow a_i+0}f(x)=b_{i+}[/tex]

[tex]f(a_i) =b_i[/tex]

Except the points of discontinuity the function is expressed by Fourier integrals whose value at the discontinuities are

[tex]f(a_i)=\frac{b_{i+} +b_{i-}}{2} [/tex]

As an example of function in post #9, Fourier integral function is f(x)=1 for any x.

- #12

Mark44

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It's been decades since I studied anything to do with topological spaces, but I don't recall anything about sequences having more than one limit. It's possible for a sequence to have subsequences that each have their own limit. For example, ##s_n = \{(-1)^n\}, n = 1, ..., \infty## is a divergent sequence, but one subsequence converges to 1, and the other converges to -1.Something else I wanted to say is that on topological spaces some sequences may have more than one limit.

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In the trivial topology of the universal set itself and the empty set all sequences converge to all points.It's been decades since I studied anything to do with topological spaces, but I don't recall anything about sequences having more than one limit. It's possible for a sequence to have subsequences that each have their own limit. For example, ##s_n = \{(-1)^n\}, n = 1, ..., \infty## is a divergent sequence, but one subsequence converges to 1, and the other converges to -1.

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S.G. Janssens

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I say it is "more useful", because the well-known characterizations of e.g. continuity and compactness in terms of sequences break down in general topological spaces, but they do hold in terms of nets.

Regarding the question about sequences having more than one limit, the extreme example is indeed as in post #13. There is a theorem that says that every net has at most one limit if and only if the topological space is Hausdorff. (Intuitively, being Hausdorff means that the space has enough open subsets.)

You can find elaborations of this in most books (with a good chapter) on point-set topology. If specific references are needed, let me know.

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Svein

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No, but a sequence can have one or moreIt's been decades since I studied anything to do with topological spaces, but I don't recall anything about sequences having more than one limit.

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S.G. Janssens

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In general topological spaces this need not be true. Post #13 is a counterexample for this, as well.A sequence has a limit iff it has one and only one cluster point.

More specifically,

##\Longrightarrow## need not be true in general topological spaces.

##\Longleftarrow## need not be true even in metric spaces.

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mathwonk

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- #18

S.G. Janssens

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Yes, it has been my understanding that Zariski topology is one of the motivations in some textbooks to

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WWGD

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In a Hausdorff topological space, sequences or nets may converge to only one limit . Otherwise, you can have as many limit points as you wish ( countably-infinite, I believe), by unioning sequences with different limits. The LimSup, LimInf of a sequence will output the largest and smallest limit points of the sequence respectively. If the two are equal to , say, L, then the sequence converges to L.

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mathwonk

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WWGD

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I don't see why sequences converge to every point on the line?

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Office_Shredder

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In a Hausdorff topological space, sequences or nets may converge to only one limit . Otherwise, you can have as many limit points as you wish ( countably-infinite, I believe), by unioning sequences with different limits. The LimSup, LimInf of a sequence will output the largest and smallest limit points of the sequence respectively. If the two are equal to , say, L, then the sequence converges to L.

I think you can get uncountably many limit points. For example let your sequence be 1/2, 1/4,2/4,3/4,1/8,2/8,...

Then every real number in [0,1] is a limit point of this sequence.

- #23

WWGD

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Then by Bolzano-Weirstrass ( Every bounded , infinite subset of Euclidean space has a limit point), the set of limit points itself would have a limit point.I think you can get uncountably many limit points. For example let your sequence be 1/2, 1/4,2/4,3/4,1/8,2/8,...

Then every real number in [0,1] is a limit point of this sequence.

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Svein

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AThen every real number in [0,1] is a limit point of this sequence.

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S.G. Janssens

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It was about sequences ofI don't see why sequences converge to every point on the line?

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Office_Shredder

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https://en.m.wikipedia.org/wiki/Limit_pointAcluster point.

a limit point (or cluster point

I think those are the same thing? Confusingly, a limit point is not the same as the limit of a sequence.

Edit: ah, perhaps I am wrong. That's the definition of a limit point of a set, and then it goes on to talk about cluster points for sequences without re-using that word.

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Svein

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https://math.libretexts.org/Bookshe...es/3.10:_Cluster_Points._Convergent_SequencesI think those are the same thing?

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Office_Shredder

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I really don't see how that contradicts the claim that in at least some contexts the word "limit point" means the same thing

https://www.emathzone.com/tutorials/real-analysis/limit-points-of-a-sequence.html#:~:text=A number l is said,values of n∈N.

So there's at least one more source that agrees with my hazy recollection. "Limit point" does not mean "Limit of the sequence", and actually means "cluster point".

At any rate, this is a matter of semantics that I think is probably not that interesting. I'm happy to agree on whatever convention for this thread.

https://www.emathzone.com/tutorials/real-analysis/limit-points-of-a-sequence.html#:~:text=A number l is said,values of n∈N.

As in the case of sets of real numbers, limit points of a sequence may also be called accumulation, cluster or condensation points.

So there's at least one more source that agrees with my hazy recollection. "Limit point" does not mean "Limit of the sequence", and actually means "cluster point".

At any rate, this is a matter of semantics that I think is probably not that interesting. I'm happy to agree on whatever convention for this thread.

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- #29

WWGD

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Thanks, I was not aware of the topology being used.It was about sequences ofdistinctpoints. (Then it is probably clear: Take any point ##x## in the space. An open nbh of ##x## is the complement of a finite set, and a sequence of distinct points leaves such a finite set eventually.)

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WWGD

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Or is ze risky ( Zariski ;) ) topology the same in the spaces in question as the finite complement topology?It was about sequences ofdistinctpoints. (Then it is probably clear: Take any point ##x## in the space. An open nbh of ##x## is the complement of a finite set, and a sequence of distinct points leaves such a finite set eventually.)

- #31

S.G. Janssens

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

- #32

WWGD

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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.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.

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S.G. Janssens

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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: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.

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mathwonk

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S.G. Janssens

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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?

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