- #1

kreil

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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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- Thread starter kreil
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- #1

kreil

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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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- #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.

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- #4

kreil

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Thanks guys, exactly the notation and process I was looking for

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