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{x$_n$} is a cauchy sequence such that for every N in natural numbers there exists m,n >= N such that x$_m$ < 0 and x$_n$ > 0. Prove lim x$_n$ = 0.

Attempt:

Since it's cauchy, it is convergent, say lim x$_n$ = L.

Suppose L > 0, then x$_m$ > 0 for all m sufficiently large. This means there exists N in natural numbers such that x$_m$ > 0 for all m >= N. This contradicts the assumption that we can find m >= N satisfying x$_m$ < 0.

lemma for x$_m$ for all m sufficiently large:

Take epsilon = L > 0, then there exists N in natural numbers such that -epsilon < x$_n$ - L < epsilon which is 0 < x$_n$ < 2L

Suppose L < 0, then x$_n$ < 0 for all n sufficiently small. This means there exists N in natural numbers such that x$_n$ < 0 for all n >= N. This contradicts the assumption that we can find n >= N satisfying x$_n$ > 0.

Since L is not less than or greater than 0, then L must equal zero.

How is this proof?

EDIT: Added proof for x$_m$ for all m sufficiently large. Is it right?

Adagio
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1 Answers1

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The idea of the proof works. There is at least one step you haven't made precise:

"Suppose $L > 0$, then $x_m > 0$ for all m sufficiently large. This means there exists N in natural numbers such that $x_m > 0$ for all $m > N$."

This is true but a completely rigorous proof would say why it is true.

Bernard W
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