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For the multivariable case, I haven't grasped the idea of open disk in the definition? Why do these definition have open disk as constituent? Why can't close disk be included?

Help will be appreciated.

thanks

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- Thread starter strugglinginmat
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- #1

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For the multivariable case, I haven't grasped the idea of open disk in the definition? Why do these definition have open disk as constituent? Why can't close disk be included?

Help will be appreciated.

thanks

- #2

morphism

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I don't really understand what you're asking in your first two questions. Can you be more explicit? What does defining the function at every point in R have to do with continuity?

As for your second question, the open disk definition is just an attempt to generalize the single variable definition. Let's take the definition of convergence of a sequence of real numbers (x_n) to some real number x: for every e>0 we can find a positive integer N such that if n>N then |x_n - x|<e. If we look more closely, the definition is saying that for any e>0, no matter how small, we can make terms of the sequence (x_n) lie in (x-e, x+e) after a certain point. Now note that on the real line an open disc is just an interval, and an interval of x will contain some 'symmetric' interval of the form (x-e, x+e) if e is chosen small enough. So we can rephrase the definition: for any open disc D containing x, we can find an N such that x_n is in D for all n>N.

As to why we choose open discs, it's just a matter of convention. Try proving that if we replace "open" with "closed" we get equivalent definitions. (Hint: inside each open disc we can fit a concentric closed disc, and vice versa.)

As for your second question, the open disk definition is just an attempt to generalize the single variable definition. Let's take the definition of convergence of a sequence of real numbers (x_n) to some real number x: for every e>0 we can find a positive integer N such that if n>N then |x_n - x|<e. If we look more closely, the definition is saying that for any e>0, no matter how small, we can make terms of the sequence (x_n) lie in (x-e, x+e) after a certain point. Now note that on the real line an open disc is just an interval, and an interval of x will contain some 'symmetric' interval of the form (x-e, x+e) if e is chosen small enough. So we can rephrase the definition: for any open disc D containing x, we can find an N such that x_n is in D for all n>N.

As to why we choose open discs, it's just a matter of convention. Try proving that if we replace "open" with "closed" we get equivalent definitions. (Hint: inside each open disc we can fit a concentric closed disc, and vice versa.)

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- #3

CompuChip

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For example, in topology, convergence is defined in terms of "neighborhoods", where a neighborhood of

which defines the concept of limit.

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mathwonk

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thus you need two differenbt concepts, one which describes the vakue at x, thats easy its just f(x). but it is harder to describe the values away from x. there are so many of them. thats where the limit definition comes in.

the limit however must be described purely in terms of values f(y) for y not equal to x, or else it wouod not be useful.

so the limit of f(y) as y approaches x, is the value thT IT LOOKS AS IF F SHIOULD HAVE AT X, MERELY FROM LOOKING AT OTHER values of f(y).

then we compare it to f(x) to see if f has the value it "should" have. continuious functions do. i.e. a continuius function is one such that you can predict f(x) by knowing f(y) for all y near x but still different from x.

in two variables you have more points near x, so you have to allow them on all sides, so you have a whole disk of them.

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The function has a limit everywhere. [tex]\lim_{x\mapsto a} f(x) =1[/tex]

However, the function does not have a value at x=0. f(0)=0/0 which is not defined. So it is possible for the limit to exist at a point even if the function has no value there.

Of course with a discontinuous function, it's possible for the limit to exist at a point and the function to be defined at the same point, but that these two values do not coincide at that point. You have to use piecewise definition for this or something like it. A regular relation won't get you this I think.

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