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The definition of continuity I am using is the following:

Let $f$ be a real function, $a\in D(f)$. If for any sequence $\{x_n\}$ in $D(f)$ converging to $a$: $$\lim f(x_n)=f(a)$$ then $f$ is continuous at the point $a$.

Now there's one detail that bothers me. The text I am using (lecture notes) then briefly mentions:

Notice that we can only consider monotonic sequences

My question is simple: why only monotonic?

Thanks.

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I suppose the text is implicitly asserting that the definition holds if and only if the restricted definition for all monotone sequences holds. Using this restricted definition probably simplifies some argument later on. So all you have to do is prove that the two definitions are equivalent. –  user108903 Jan 15 '13 at 8:35

2 Answers 2

up vote 2 down vote accepted

Assuming the domain of the function is a subset of $\mathbb R$ (otherwise it is not clear what monotonous should mean) I think the remark should be "Notice that it suffices to consider only monotone sequences". Clearly, if the condition holds for any sequence that it holds for a monotone one and so the the more general condition implies the one for monotone sequences. On the other hand, any sequence in $\mathbb R$ contains a monotone subsequence and you can use this to show that the more restrictive condition implies the more general one. Thus they are equivalent.

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Aha! I get it now. It's more of a linguistic problem, actually. In Czech the word "lze" can be understood as both "can only be" and "it is possible to" and the second possibility didn't occur to me. Thanks for clearing that up! –  Dahn Jahn Jan 15 '13 at 8:39
    
you're welcome. –  Ittay Weiss Jan 15 '13 at 8:40

Because you can always take a monotone sequence from an arbitrary sequence. So if for some sequence ${x_n}$ in D(f) converging to a : $limf(x_n )$ not converges to f(a), then there is a subsequence $y_n$ in $x_n$, such that $|f(y_n)-f(a)|$>r>0, then take a monotone subsequence of ${y_n}$ you get that there is a monotone sequence such that the condition is not satisfied, this shows that you just need to consider the monotone sequence.

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Thanks for the answer. –  Dahn Jahn Jan 15 '13 at 8:40
    
you are welcome. –  lee Jan 15 '13 at 8:42

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