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Discontinuity

Working through some problems and stuck on this one.



Managed definition:

"There exists epsilon > 0 such that for all delta > 0, |x - x_0| <= delta & |f(x) - f(x_0) => epsilon."

However, stuck on the second part. What I've done so far:

"Choose x_n such that |x_n - x_0| <= 1/n and let delta = 1/n.

|x_n - x_0| <= 1/n implies x_n - x_0 -> 0 as n -> infinity implies x_n -> x_0 as n -> infinity.

Assume f(x_n) -> f(x_0) as n -> infinity. However, by definition of discontinuity, this is a contradiction."

My solution feels a little flimsy, especially the second part where I just state it cannot be true. Any feedback will be appreciated.

Regards,
19292
WP_20141110_002.jpg59.5KB
Original post by 19292
Working through some problems and stuck on this one.



Managed definition:

"There exists epsilon > 0 such that for all delta > 0, |x - x_0| <= delta & |f(x) - f(x_0) => epsilon."There are some words missing here, E.g.

""There exists epsilon > 0 such that for all delta > 0 we can find x s.t. , |x - x_0| <= delta & |f(x) - f(x_0) => epsilon."

(If I were examining I'd give you the benefit of the doubt on this, except that your subsequent working doesn't convince me that you really knew what you meant to say...

However, stuck on the second part. What I've done so far:

"Choose x_n such that |x_n - x_0| <= 1/n and let delta = 1/n.

|x_n - x_0| <= 1/n implies x_n - x_0 -> 0 as n -> infinity implies x_n -> x_0 as n -> infinity.

Assume f(x_n) -> f(x_0) as n -> infinity. However, by definition of discontinuity, this is a contradiction."

My solution feels a little flimsy
Yes, this is very flimsy.

The key point here is that by the definition of discontinuiuty, we can take δ<1/n\delta < 1/n and find x_n s.t. ∣xn−x0∣<δ|x_n - x_0| < \delta but ∣f(xn)−f(x0)∣>ϵ|f(x_n)-f(x_0)| > \epsilon. (*)

But if f(x_n) -> f(x_0), we must have |f(x_n) - f(x_0)| -> 0, contradicting (*).
Reply 2
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19292
OP
Thanks for the constructive feedback.

Original post by DFranklin
δ<1/n\delta < 1/n


However, this bit confuses me. I can see how this can imply |x_n - x_0| < delta if delta = 1/n but not with the sign this way around.

Thanks again.

Regards,
19292
(edited 9 years ago)
Original post by 19292
]However, this bit confuses me. I can see how this can imply |x_n - x_0| < delta if delta = 1/n but not with the sign this way around.We are not "implying" that |x_n - x_0| is < delta.

By the definition we (I) provided:

"...for all delta > 0 we can find x s.t. , |x - x_0| <= delta & |f(x) - f(x_0) => epsilon."

So |x_n - x_0| is <= delta because we can find x_n s.t. |x_n - x_0|<=delta.
Reply 4
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19292
OP
Original post by DFranklin
We are not "implying" that |x_n - x_0| is < delta.

By the definition we (I) provided:

"...for all delta > 0 we can find x s.t. , |x - x_0| <= delta & |f(x) - f(x_0) => epsilon."

So |x_n - x_0| is <= delta because we can find x_n s.t. |x_n - x_0|<=delta.


That makes sense now. Feel silly I didn't get it before. Thanks a lot. Much appreciated.

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