- #1

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is this definition equivalent to the epsilon-delta definition?

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- Thread starter JanEnClaesen
- Start date

- #1

- 59

- 4

is this definition equivalent to the epsilon-delta definition?

- #2

CompuChip

Science Advisor

Homework Helper

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Yes, it is almost a literal translation. Try writing out the limit definition, then note that if ##\epsilon > 0##, "for all x such that ##|x - a| < \epsilon##" means "for all x in ##[a - \epsilon, a + \epsilon]##"; and ##x \in A \implies x \in B## means that "B is a subset of A", in other words, "A includes B".

- #3

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Under your definition,

- #4

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The problem f([a-d, a+d]) certainly fixes is that it requires all the points in the interval to be part of the range.

- #5

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... so that if we use this definition of continuity every symmetric function is continuous around its centre of symmetry a.

[itex]f([a-\delta,a+\delta])[/itex] is only a singleton if [itex]f[/itex] is constant on the whole interval [itex][a-\delta,a+\delta][/itex], a

- #6

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[itex]f([a-\delta,a+\delta])[/itex] is only a singleton if [itex]f[/itex] is constant on the whole interval [itex][a-\delta,a+\delta][/itex], amuchstronger restriction than just [itex]f(a-\delta)=f(a)=f(a+\delta)[/itex].

You're right, I made an error of interpretation.

- #7

pasmith

Homework Helper

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Hi Jan,

Yes, it is almost a literal translation. Try writing out the limit definition, then note that if ##\epsilon > 0##, "for all x such that ##|x - a| < \epsilon##" means "for all x in ##[a - \epsilon, a + \epsilon]##";

The intervals must be open: [itex](a - \epsilon, a + \epsilon)[/itex].

- #8

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But the given definition with closed neighbourhoods is still equivalent.

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