Imagine you are a child or an AI robot with an incredible intelligence. You become fascinated and amused by informally thinking about (with no references) the finite symmetric groups $S_n$. Eventually you want to formalize this 'slice of math', and attempt to layout a formal theory. You already understand how to construct the finite von Neumann ordinals,

0   = {}           = ∅
1   = {0}          = {∅}
2   = {0, 1}       = {∅, {∅}}
3   = {0, 1, 2}    = {∅, {∅}, {∅, {∅}}}
4   = {0, 1, 2, 3} = {∅, {∅}, {∅, {∅}}, {∅, {∅}, {∅, {∅}}}}
etc.

and regard these sets as canonical.

You decide that each of these collections of automorphisms must have an identity and begin by explicitly constructing $S_1$. Using recursion, you know that with $S_n$ defined you can construct $S_{\sigma(n)}$ where $\sigma(n)$ is the next ordinal.

So you've constructed a chain of proper natural inclusions,

$\quad S_1 \hookrightarrow S_2 \hookrightarrow S_3 \hookrightarrow \dots $

You develop your theory further and note that

$\;$ There is one and only one group structure on a singleton set.

$\;$ There is one and only one homomorphism of $S_1$ into $S_n$.

$\;$ There is one and only one homomorphism of $S_n$ into $S_1$.

Just for fun you decide to postulate the following as an axiom,

$\; \text{There exist a group } S_\omega \text{ such that for every } x \in S_\omega \text{ there exists an ordinal } n \text{ with } x \in S_n$

finding no contradictions and concluding that $S_\omega$ must be unique.

You also observe that there is one and only one way to re-frame a singleton set as a pointed set.

Having studied philosophy, you recall the quote

A journey of a thousand miles must begin with a single step.

Lao Tzu

A terminal object in a category by definition is an object T with the property that for any object C in the category, there is exactly one arrow C -> T. In the category of sets, a singleton object, for example {1} has this property:

For any set S there is a function f:S -> {1} defined by setting f(s)=1 for any element s of S. There can't be another function g:S -> {1}, because for some element s of S, g(s) would have to be an element t of {1} other than 1, but there is no such element t. So there is exactly one function from S to {1} for each object S of the category.

Your remark about there being many functions from {1} to a set S is irrelevant; they go in the wrong direction to be relevant to the definition of terminal object.

A set is a mathematical object; a singleton set is a set with only one element.

The set N of natural numbers has infinite many elements.

The singleton set { N } has only one element.

In general, the properties of an object and those of the set with that object as single elements are not the same.

As per example above, the singleton { N } has only one elements while its (only) element have infinitely many element.

But we may consider the singleton { emptyset }; again, it has one element, while its element has no elements.

As has been noted, the notion of "open" and "closed" is not absolute, but depends on a topology. So in order to answer your question one must first ask what topology you are considering.

A topological space is a pair, $(X,\tau)$, where $X$ is a nonempty set, and $\tau$ is a collection of subsets of $X$ such that:

• $\emptyset$ and $X$ are both elements of $\tau$;
• If $A$ and $B$ are elements of $\tau$, then $A\cap B$ is an element of $\tau$;
• If $\{A_i\}_{i\in I}$ is an arbitrary family of elements of $\tau$, then $\bigcup_{i\in I}A_i$ is an element of $\tau$.

The elements of $\tau$ are said to be "open" (in $X$, in the topology $\tau$), and a set $C\subseteq X$ is said to be "closed" if and only if $X-C\in\tau$ (that is, if the complement is open).

In $\mathbb{R}$, we can let $\tau$ be the collection of all subsets that are unions of open intervals; equivalently, a set $\mathcal{O}\subseteq\mathbb{R}$ is open if and only if for every $x\in\mathcal{O}$ there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq\mathcal{O}$. You may want to convince yourself that the collection of all such sets satisfies the three conditions above, and hence makes $\mathbb{R}$ a topological space. This topology is what is called the "usual" (or "metric") topology on $\mathbb{R}$.

If you are working inside of $\mathbb{R}$ with this topology, then singletons $\{x\}$ are certainly closed, because their complements are open: given any $a\in \mathbb{R}-\{x\}$, let $\epsilon=|a-x|$. Then $x\notin (a-\epsilon,a+\epsilon)$, so $(a-\epsilon,a+\epsilon)\subseteq \mathbb{R}-\{x\}$; hence $\mathbb{R}-\{x\}$ is open, so $\{x\}$ is closed.

The reason you give for $\{x\}$ to be open does not really make sense. Every set is a subset of itself, so if that argument were valid, every set would always be "open"; but we know this is not the case in every topological space (certainly not in $\mathbb{R}$ with the "usual topology"). So that argument certainly does not work.

So: is $\{x\}$ open in $\mathbb{R}$ in the usual topology? Well, $x\in\{x\}$. Does there exist an $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq \{x\}$? If so, then congratulations, you have shown the set is open. If there is no such $\epsilon$, and you prove that, then congratulations, you have shown that $\{x\}$ is not open.