- #1
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could anyone produce a proof showing why a sequence can't have more than 1 unique limit?
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Sequence can't have more than 1 unique limit?
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Discussion Overview
The discussion centers on the question of whether a sequence can have more than one unique limit, exploring proofs and reasoning related to this concept in mathematical analysis.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant requests a proof demonstrating that a sequence cannot have more than one unique limit.
- Another participant presents a proof using the definition of limits, arguing that if a sequence has two limits, it leads to a contradiction, concluding that the two limits must be equal.
- A third participant offers a proof by contradiction, stating that if a sequence converges to two different limits, the intervals around those limits must be disjoint, which leads to a contradiction regarding the sequence's terms.
- This participant expresses a preference for the visual or geometric interpretation of convergence over traditional epsilon-delta arguments.
Areas of Agreement / Disagreement
Participants generally agree on the proofs presented that support the idea that a sequence cannot have more than one unique limit, but the discussion does not explicitly resolve any potential disagreements about the methods or interpretations of the proofs.
Contextual Notes
The discussion relies on the definitions of limits and convergence, and the proofs assume familiarity with epsilon-delta arguments and the properties of intervals. There is no exploration of alternative definitions or contexts where the question might differ.
- #2
quasar987
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Suppose {x_n} has two limits, say x and y. Then this means that
1) for all e > 0, there exists and N_1 > 0 s.t. n > N_1 ==> |x_n - x| < e
2) for all e > 0, there exists and N_2 > 0 s.t. n > N_2 ==> |x_n - y| < e
It follows that for n > max{N_1, N_2}, |x_n - x| < e and |x_n - y| < e.
Therefor, for n > max{N_1, N_2}, |x - x_n + x_n - y| [itex]\leq[/itex] |x - x_n| + |x_n - y| = |x - x_n| + |x_n - y| < e + e = 2e.
But |x - x_n + x_n - y| = |x - y|. So this last inequality writes, in limit notation,
[tex]\lim_{n \rightarrow \infty} (x - y) = 0[/tex]
but x - y = a constant. We know that the limit of a constant is the constant itself. So [itex]\lim_{n \rightarrow \infty} (x - y) = x - y[/itex]. So x - y = 0.
Edit: So x = y. [itex]\blacksquare[/itex] (P.S. No, I don't think you're that dumb, I just like my proofs complete to the last drop... kind of a obsesso-maniacal thing I have
)
1) for all e > 0, there exists and N_1 > 0 s.t. n > N_1 ==> |x_n - x| < e
2) for all e > 0, there exists and N_2 > 0 s.t. n > N_2 ==> |x_n - y| < e
It follows that for n > max{N_1, N_2}, |x_n - x| < e and |x_n - y| < e.
Therefor, for n > max{N_1, N_2}, |x - x_n + x_n - y| [itex]\leq[/itex] |x - x_n| + |x_n - y| = |x - x_n| + |x_n - y| < e + e = 2e.
But |x - x_n + x_n - y| = |x - y|. So this last inequality writes, in limit notation,
[tex]\lim_{n \rightarrow \infty} (x - y) = 0[/tex]
but x - y = a constant. We know that the limit of a constant is the constant itself. So [itex]\lim_{n \rightarrow \infty} (x - y) = x - y[/itex]. So x - y = 0.
Edit: So x = y. [itex]\blacksquare[/itex] (P.S. No, I don't think you're that dumb, I just like my proofs complete to the last drop... kind of a obsesso-maniacal thing I have
)
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- #3
matt grime
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Another one, by contradiction (which is essentially the same as the previous one) is: if x_n tends to x and x_n tends to y. If x=/=y, then let e=|x-y|/3, then for all n sufficently large we have
|x_n - x| < e , ie all the x_n are in the interval (x-e,x+e) for n >N
and
|x_n - y| <e, ie all the x_n are in the interval (y-e,y+e) for n>N
but the two intervals (x-e,x+e) and (y-e,y+e) are disjoint so it is impossible for all the x_n for n>N to be in two disjoint intervals. a contradiction. the only assumption was that x=/=y, so it must be that x=y.
Personally I prefer this one (even though it is an unnecessary proof by contradiction) because it makes you think visually (or geometrically) about what it means for a seqqunce to converge: for all open intervals about x (or open "balls" in higher dimensions) that if we throw away some finite number of terms at the start of the sequence then all of the ones that are left are inside the inteval or ball. it gets awayy from the epsilon delta arguments, and that has to be a good thing.
|x_n - x| < e , ie all the x_n are in the interval (x-e,x+e) for n >N
and
|x_n - y| <e, ie all the x_n are in the interval (y-e,y+e) for n>N
but the two intervals (x-e,x+e) and (y-e,y+e) are disjoint so it is impossible for all the x_n for n>N to be in two disjoint intervals. a contradiction. the only assumption was that x=/=y, so it must be that x=y.
Personally I prefer this one (even though it is an unnecessary proof by contradiction) because it makes you think visually (or geometrically) about what it means for a seqqunce to converge: for all open intervals about x (or open "balls" in higher dimensions) that if we throw away some finite number of terms at the start of the sequence then all of the ones that are left are inside the inteval or ball. it gets awayy from the epsilon delta arguments, and that has to be a good thing.
Last edited:
- #4
- 665
- 68
Thanks guys, exactly the notation and process I was looking for
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