# Difference between a function being continuos at at point and havinga limit there

1. Oct 29, 2007

### 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... ?

Last edited: Oct 29, 2007
2. Oct 29, 2007

### NateTG

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

3. Oct 29, 2007

### Smartass

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

4. Oct 29, 2007

### zhentil

A function is vacuously continuous at isolated points.

5. Oct 29, 2007

### futurebird

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???

6. Oct 30, 2007

### 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

Last edited: Oct 30, 2007
7. Oct 30, 2007

### zhentil

Yes. $$|x-7| < 1/2 \Rightarrow |f(x)-f(7)| = 0.$$

Last edited: Oct 30, 2007