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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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Stephen Tashi

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

- #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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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.

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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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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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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,...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.

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

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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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