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

Question

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 $\forall \epsilon > 0 \; \exists \delta > 0 \; \forall x \; [0<|x-a|<\delta\implies |f(x)-l|<\epsilon]$

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

Answer 1

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 $$\forall\epsilon>0,\exists\delta>0:\forall x\color{red}{\in\operatorname{dom}f},\:\bigl[0<|x-a|<\delta\implies |f(x)-l|<\epsilon\bigr]$$ Note that this definition also varies subtly and critically from Spivak's.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

Spivak's definition doesn't actually break for functions like yours, like the accepted answer says, due to the logical subtlety (annoyance?) of vacuous truths, which comes from the definition (truth table) of the logical operator (or logical connective, I never know the difference) of logical implication.

Namely, in zeroth-order logic ("propositional logic"), an implication logical statement of the form

$$A \implies B$$

is false iff $A$ is true and $B$ is false; in particular, it is true if the hypothesis $A$ is false, regardless of the truth value of $B$.

In first-order logic ("predicate logic"), an implication logical statement of the form

$$\forall x \left< A(x) \implies B(x) \right>$$

(where $A(x)$ is a logical formula whose truth value depends on the value of the variable $x$, and likewise for $B(x)$) is false iff, for all $x$, $A(x)$ is true and $B(x)$ is false; in particular, it is true if, for all $x$, the hypothesis $A(x)$ is false, regardless of the truth value of $B(x)$. (For nested quantifiers like $\forall\exists\forall$ in the limit definition, the truth value "bubbles up" from the innermost levels to the outermost levels, I think).

In the case of Spivak's limits and your function, the implication logical statement

$$\forall x \in \text{Dom}(f) \left< 0 < |x - a | < \delta \implies |f(x) - f(a)| < \epsilon \right>$$

is true because, for all $x \in \text{Dom}(f)$, the hypothesis $0 < |x - a| < \delta$ is false (because $\text{Dom}(f)$ has single element, namely $1$, and the statement $0 < |1 - 1| < \delta$ is false regardless of the value of $\delta$).

Now imma handwave and say that the entire formula (which is a formula in 2 variables, namely the function $f$ and the number $a$)

$$\forall \epsilon > 0\ \ \exists \delta >0\ \ \forall x \in \text{Dom}(f) \left< 0 < |x - a | < \delta \implies |f(x) - f(a)| < \epsilon \right>$$

is true when you set $f$ to your function and $a$ to be $1 \in \text{Dom}(f)$.

So, by the Spivak definition of limit, the limit of $f$ at $1$ is $1$.

And, by the definition of $f$, $f$ maps 1 to 1.

So, by the Spivak definition of continuity, $f$ is continuous at $1$.

(Disclaimer: I'm not sure of anything I just said.)

(The real complication is not the "annoyance" of the implication logical operator, but the fact that people use it in the first place, and for virtually every statement in math. Why not use the and operator, or the iff operator, instead? I suppose there's a reason, but I don't know what it is. Notice that the implication logical operator is related to the geometry/topology of subsets, while the and logical operator and the iff logical operator are related to the geometry/topology of intersections. (Look at the Venn diagrams.))