Can a function with just one point in its domain be continuous?

For example if my function is $f:\{1\}\longrightarrow \mathbb{R}$ such that $f(1)=1$.

I have the next context:

1) According to the definition given in Spivak's book and also in wikipedia, since $\lim_{x\to1}f$ doesn't exist because $1$ is not an accumulation point, then the function is not continuous at $1$ (Otherwise it should be $\lim_{x\to 1}f=f(1)$).

2) According to this answer , as far as I can understand a function is continuous at an isolated point.

I don't understand.

Edit:

• Spivak's definition of limit: The function $f$ approaches to $l$ near $a$ means $\forall \epsilon >0\exists \delta >0\forall x (0<|x-a|<\delta\longrightarrow |f(x)-l)|<\epsilon)$

• Spivak's definition of continuity: The function $f$ is continuous at $a$ if $\lim_{x\to a}f(x)=f(a)$

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Your claim in 1) that $lim_{x\to 1} f(x)$ does not exist because $1$ is not a accumulate point is wrong. If you do not have any sequence converging to $1$ which is different of $1$, then automatically the definition of continuity by sequence is satisfied. –  Tomás Jul 30 '13 at 1:20
Any function is continuous on its isolated points. You can easily argue this from the $\varepsilon$-$\delta$ definition. –  Cameron Williams Jul 30 '13 at 1:22
I don't know Spivak's but I have seem some books in which they do what you are saying. They define continuity as a property that only applies to functions and a point that is a cluster point of the domain. Strictly speaking you are right that in that case continuous would be an adjective that doesn't apply to a function as in your case. Now, you can see from the answers that such definition is not the most general and is only used in the more basic books, in which only very basic functions and problems are studied. –  ABC Jul 30 '13 at 1:37

Based on the definitions Spivak gave, I suspect that (as discussed in comments) his definition of continuity is based on the assumption that we're dealing with functions defined everywhere, or at very least having domains with no isolated points. His definition does indeed break down (badly) for functions such as yours.

A related (but more general) definition given for continuity at a point $a$ of the domain of a function $f$ is something like $$\forall\epsilon>0\:\exists\delta>0\:\forall x\in\operatorname{dom}f\:\bigl[|x-a|<\delta\implies |f(x)-f(a)|<\epsilon\bigr]$$ This is provably equivalent to:

(i) $x$ is isolated in $\operatorname{dom}f$, or

(ii) $x$ is a point of accumulation of $\operatorname{dom}(f)$ and $\lim_{y\to x}f(y)=f(x)$.

The key to the proof is that for a point of accumulation $a$ of $\operatorname{dom}f,$ we say $\lim_{x\to a}f(x)=l$ iff $$\forall\epsilon>0\:\exists\delta>0\:\forall x\in\operatorname{dom}f\:\bigl[\color{red}{0<}|x-a|<\delta\implies |f(x)-l|<\epsilon\bigr]$$ Note that this definition also varies subtly and critically from Spivak's.

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The definition in Spivak's book is just that $f$ is continuous at $x$ if and only if $\lim_{y\to x}f(y)=f(x)$ having defined limit just the same as you did in your last statement. But defining $f$ to be continuous in this way doesn't imply the definition in your first statement, which seems to be more general. Then I guess the definition of Spivak is somehow restricted so it's not that my function given above is not continuous at $1$ but rather there's no sense to ask that because $1$ is not an accumulation point and then the definition of limit doesn't apply. Does this make sense? –  Daniela Diaz Jul 30 '13 at 4:29
What I mean is that in general sense it is not always that $f$ is continous at $x$ if and only if $\lim_{y\to x}f(y)=f(x)$. Am I right? –  Daniela Diaz Jul 30 '13 at 4:33
True. It seems that Spivak's definition is that of continuity at an accumulation point of the domain. When dealing with the more basic functions of introductory math courses, you won't encounter any isolated points of function domains, so for such basic functions, the definitions agree. That's probably why Spivak chose that definition. –  Cameron Buie Jul 30 '13 at 4:42

You can also look at the general, topological definition of continuity in terms of open sets; namely, $f$ is continuous means that for all open set $O$ of $\mathbb R$, $f^{-1}(O)$ is an open of $\{1\}$. Since $f^{-1}(O)$ is either $\emptyset$ or $\{1\}$, and both are open sets (for the topology on $\{1\}$), $f$ is continuous.

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If you take $\lim_{x \rightarrow a} f(x) = f(a)$ to be the definition of continuity, and if what we mean by the limit is that for all $\epsilon >0$ there is a $\delta >0$ such that $x\in D$, $D$ the domain of $f$, and $0<|x-a|<\delta$, we have $|f(a)-f(x)|<\epsilon$, then yes, the limit is ill-defined.

If, on the other hand, you take continuity to be for all $\epsilon >0$ there is a $\delta>0$ such that $x\in D$ and $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$, then we clearly have continuity in a one point domain.

EDIT: I don't know if you are aware of much topology, but one can show using the appropriate definition of continuity that it is equivalently to the following: A function $f:A\rightarrow R$ is continuous iff for all open sets $U \subseteq R$ the preimage $f^{-1}(U)\subseteq A$ is open with respect to the topology on $A$.

Suppose we start with this definition instead. If we take the usual induced topology on a single point set $A=\{a\}$, then of course $A$ is open. Thus, $f^{-1}(U)$ is either the empty set or $A$ for any open set $U$. Continuity then follows.

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I think that it should not matter whether you look at a punctured neighborhood or a regular neighborhood, since in the case that you have a punctured neighborhood you are are performing a universal quantification over the empty set. I would want to look at the exact quantifies however. –  Baby Dragon Jul 30 '13 at 2:36
+1 I think that saying that limit is ill-defined is not very proper because this definiton is the one given in many calculus books. I think it's just not as general as the one given in books of topology. Anyway your answer made me think a lot about it, thanks. –  Daniela Diaz Jul 30 '13 at 4:55
@BabyDragon: You're absolutely correct about universal quantification over the empty set. The problem there becomes that if $a$ is isolated in the domain of $f$, then we can replace $f(a)$ with any real $y$ in the $\epsilon$-$\delta$ definition of the limit of $f(x)$ as $x$ approaches $a$--we lose uniqueness (badly), so the limit is indeed ill-defined. –  Cameron Buie Jul 30 '13 at 10:31
@DanielaDiaz: You'll probably also be interested in my comment to Baby Dragon. –  Cameron Buie Jul 30 '13 at 10:58
@DanielaDiaz: Yes, indeed! And looking at your updated post, Spivak seems to be further assuming that we'll be dealing only with functions defined everywhere (note the $\forall x$, rather than $\forall x\in\operatorname{dom}f$). I will update my answer accordingly. –  Cameron Buie Jul 30 '13 at 11:24

In C.H. Edward's Advanced Calculus text an point in a set $S$ which is contained in the center of some open ball which contains no other points in $S$ is called an isolated point. The definition of continuous functions includes a comment that functions are considered continuous at isolated points by default. Of course, technically, isolated points are not limit points so this case will be lost in some other discourse. A nice result of this convention is that functions with discrete domain are by default continuous. For example, sequences are continuous.

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