A function whose domain is a singleton is always continuous . How? According to a text,
Suppose f is a function defined on aclosed interval [a, b], then for f to be continuous, it needs to be continuous at everypoint in [a, b] including the end points a and b. Continuity of f at a means the right hand limit and left hand limit of f at a and b respectively equal value of f at a and b. Because the left and right hand limit of f at a and b have no meaning. As a consequenceof this definition, if f is defined only at one point, it is continuous there, i.e., if thedomain of f is a singleton, f is a continuous function.
How does the fact that function whose domain is singleton follow from above definition? Can someone explain and please dont use the epsilon delta definition of limit use the basic or old definition or just explain theoretically.
 A: abstract mathematical answer: The most fundamental definition of continuity is "the preimage of open sets is open". And a singleton set has trivial topology so there is nothing non-open there.
intuitive answer (hopefully): continuity of a function $y=f(x)$ at a point $x$ means:
IF you approach $x$ in any way (either from right, left, or somehow jumping around),
THEN you will always also approach the correct $y$ value.
But if the domain of your function is only a single point, it is impossible to approach $x$ from anywhere. Therefore the "IF" part of the condition can never happen, so there is no condition to check. Therefore any such function is continuous.
A: I was wondering myself in a simple cohomology problem and after looking at this post, I came up with a different answer from Simon:
The short answer is that $f$ is constant and so is continuous.


*

*The problem with this is that the rule of constant map being continuous is that for $f: A \to \{b\}$, we have $f$ continuous, but we don't exactly have an explicit rule (at least in my textbook. Probably another textbook has the following explicit rule) that says for $f: A \to B$ and the range of $f$, $f(A)$, is a singleton, we have $f$ continuous. I prove as follows.


The long answer: Constant maps, inclusion maps and compositions are continuous.


*

*Let $f: \{a\} \to Y$, for a "topological space" $Y$ (For calculus: take for granted that any interval in $\mathbb R$, including singletons, $\mathbb R$ and $\emptyset$, are "topological spaces"). Let $\tilde{f}:\{a\} \to \{f(a)\} = \{\tilde{f}(a)\}$ be $f$ with restricted range, i.e. $\iota \circ \tilde{f} = f$ for inclusion map $\iota$ (see A1).

*Constant maps are continuous.

*By (1), $\tilde{f}$ is constant.

*By (2), $\tilde{f}$ is continuous.

*Inclusion maps from any 2 "topological spaces" (Again, for calculus: take for granted that any interval in $\mathbb R$, including singletons, $\mathbb R$ and $\emptyset$, are "topological spaces") are always continuous.

*Compositions of continuous maps are continuous.

*The $\iota$ in (1) is continuous by (5).

*$f$ is continuous by (1), (5) and (6).

(A1) This means $\iota(b)=b, \iota: \{f(a)\} \to Y, b \in \{f(a)\}$.
Inclusion maps are just fancy ways to distinguish the maps like $f: \mathbb R \to [0,\infty), f(x)=x^2$ and $g: \mathbb R \to \mathbb R, g(x)=x^2$. What's the difference between $f$ and $g$? From an engineering perspective, nothing. As machines that give a single output any single input, $f$ and $g$ are identical. From a mathematical perspective, not nothing but not much. As mathematical objects, $f$ and $g$ have different ranges and are therefore different. However, they are related by $g = \iota \circ f$ for the inclusion map $\iota: [0,\infty) \to \mathbb R$
Thus, if you ignore all the fancy stuff, such as inclusions and topological spaces, then you can have the short answer.
