Understanding the Difference: Continuous vs Limit at Point

[futurebird]

I'm confused about a point that my book on real analysis is making about the difference between the definition of a function being continuos at at point and the definition of a function having a limit L at a point.

My rough understanding of the matter is that in order for a function to be continuous at a point the function must have a limit L at the point (from both sides) and the values of the function at that point must be L. However it is possible to have a limit at a point, even if the value of that function at the point is not the same as the limit.

From the book:

DEFINITION: Suppose E is a subset of R and f: E \rightarrow R. If x_0 \in E, then f is continuous at x_0 iff for each \epsilon > 0, there is a \delta > 0 such that if

|x-x_0|< \delta, with x \in E,

then

|f(x)-f(x0)|< \epsilon.

Compare this with the definition of the limit of a function at a point x_0 . First of all, for continuity at x_0, the number must belong to E but it need not be an accumulation point of E. ( Is this saying that f(x) = L? If not what is it saying? ) Indeed, if f: E \rightarrow R with x_0 \in E and x_0 not an accumulation point of E, then there is a \delta > 0 such that if |x - x_0 | < \delta and x \in E, then x=x_0 .​

That last bit has me lost. Why would x = x_0 if x_0 is not an accumulation point of E?

I think that x_0 , not being an accumulation point means that there is at least one neighborhood of x_0 that contains a finite number, possibly zero vales of f(x) when x is in a delta neighborhood of x_0. But how does this force x = x_0?

The book defines accumulation points interms of series:

A real number A is an accumulation point of S iff every neighborhood of A contains infinitely many points of S.​

So, not an accumulation point would mean that every neighborhood of A contains finite points of S, or zero ponits of S... ?

 

[NateTG]

I think that x_0 , not being an accumulation point means that there is at least one neighborhood of x_0 that contains a finite number, possibly zero vales of f(x) when x is in a delta neighborhood of x_0. But how does this force x = x_0?

Let's say that x_0 \in E is not an accumulation point of E. Then there is some \delta so that:
E \cap (x_0 - \delta,x_0+\delta) = x_0[/itex]<br /> <br /> Then the hypothesis:<br /> |x-x_0|&amp;lt;\delta<br /> and<br /> x \in E<br /> is equivalent to<br /> x_0 \in E \cap (x_0 - \delta,x_0+\delta) = x_0<br /> which implies<br /> x_0=x<br /> (and obviously)<br /> f(x_0)-f(x)=0

 

[Smartass]

...
is equivalent to
x_0 \in E \cap (x_0 - \delta,x_0+\delta) = x_0
...

Shouldn't it be x \in E \cap (x_0 - \delta,x_0+\delta) = x_0?

 

[zhentil]

A function is vacuously continuous at isolated points.

 

[futurebird]

A function is vacuously continuous at isolated points.

That helps a lot. But let me see if I know what you mean. If I have the set of points N, the natural numbers then it's vacuously continuous a say 7?

That's odd.

Is that what this is all about?

 

[ZioX]

The point is trivial. x_0 an accumulation pt in E iff every nbd of x_0 contains a point in E iff every neighbourhood of x_0 contains infinitely many points in E. If x_0 is not an accumulation pt in E then there exists a nbd of x_0 such that it only contains finitely many points in E.

Trivial point: Well, if x_0 is not an accumulation point in E then there are finitely many x's in E within some delta>0 distance from x_0. Take the closest point to x_0, this exists as we're taking the minimum of some finite set. Call the distance between x_0 and this closest point delta_min. So the only points in E with the property that |x-x_0|<delta_min is the one point x_0. (Draw a picture: this immediately clear).

Some authors absolutely refuse to talk about limits of functions at non accumulation points. The author of whatever you're quoting is obviously not one of them. The problem arises when you look at the difference between continuity and limits. Limits talk only about the elements getting closer to some point (but not the point itself), continuity talks about elements getting closer to some point AND the point itself. (0<|x-x_0|<delta => |f(x)-f(x_0)|<epsilon for limits and |x-x_0|<delta => |f(x)-f(x_0)|<epsilon for continuity)

Saying that functions are vacuously continuous at non accumulation is awkward as the limits NEVER exist, yet they will always satisfy the latter criterion for continuity above.

Take your example. The integers with the identity function. What's the limit of id(x) as x->7? If it did exist then for every epsilon there would be a delta such that 0<|x-7|<delta (x an integer) =>|x-7|<epsilon. Take epsilon=1/2. It's clear that |x-7|<1/2 and x integral iff x=7. But x cannot be 7, as then |x-7|=|7-7|=0. Hence id(x) does not have a limit as x->7. But it does satisfy the delta-epsilon definition of continuity at x=7: since 7 is not an accumulation point (only one point in any ball centered at 7 with radius strictly less than 1) we can always pick delta<1 and so |x-7|<delta<1=>x=7=>|id(7)-id(7)|=0<epsilon for any epsilon>0.

Do you see how continuity at non accumulation points is weird? Heuristically, we think of continuity as getting closer and closer to some point. But if we're talking about an non accumulation point you can't get closer and closer! There's just one lone point. Non-accumulation points are called isolated points -- this should help you remember the definition.

Summary:

f(x)->f(x_0) as x->x_0 is equivalent to continuous at x_0 (epsilon/delta defn) if x_0 is an accumulation point

 

[zhentil]

That helps a lot. But let me see if I know what you mean. If I have the set of points N, the natural numbers then it's vacuously continuous a say 7?

That's odd.

Is that what this is all about?

Yes. |x-7| &lt; 1/2 \Rightarrow |f(x)-f(7)| = 0.