The set defined as $$F=\{\langle\varnothing,a \rangle, \langle \{\varnothing\},b\rangle\}$$ a function?
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Is $\langle a,b\rangle$ notation for ordered pair? – Arturo Magidin Jun 15 '20 at 00:50
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1It may also depend on what definition of function you are using. In some definitions, you need to specify two sets in addition to a collection of ordered pairs. – Arturo Magidin Jun 15 '20 at 00:52
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Yes, $\langle a, b \rangle$ is ordered set. I think it is equivalent to ${ a, {a,b}}$? – sharkbear Jun 16 '20 at 14:28
3 Answers
Assuming you're using the "functions-as-sets-of-ordered-pairs" approach, then yes, it is: its domain is $\{\emptyset,\{\emptyset\}\}$, it sends $\emptyset$ to $a$, and it sends $\{\emptyset\}$ to $b$. Crucially $\emptyset\not=\{\emptyset\}$, so there's no inconsistency here.
(The other standard approach to functions views a function as a set of ordered pairs together with a specific mention of domain and codomain. Of course the domain is determined by the set of ordered pairs (at least, if we don't allow partial functions), but the codomain is not, so this is a genuinely different approach. If we take this stance then what you've written is not a function.)
EDIT: note that the parenthetical remark is exactly the point Arturo Magidin makes in his comment above.

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Yes: it is a function that maps $F: \{\emptyset, \{\emptyset\}\} \to \{a,b\}$.
As $\emptyset$ and $\{\emptyset\}$ are two distinct and different mathematical objects and each element of the domain, $\{\emptyset, \{\emptyset\}\}$, is mapped to precisely one element of the codomain, $\{a,b\}$ and not element of $\{\emptyset, \{\emptyset\}\}$ is mapped to more than one element of $\{a,b\}$, $F$ satisfies every definition of a function.
(Edit: Noah and Artruro make the valid point that the the domain can be inferred to be $\{\emptyset, \{\emptyset\}\}$ but the codomain is unsepecified. The codomain $\supset \{a,b\}$ and if your texts requires a codomain must be specified then this technically is not a function but that's fairly picky. For must practical purposes I believe most would consider this a function.)

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Yes, it is a function, and if you adopt the usual definition that $0 = \varnothing$, $1 = \{ \varnothing \}$, then this function maps from $\{ 0, 1 \}$ to real (or complex) numbers, and its explicit form is: $F(0) = a, F(1) = b$.

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1This is a good short answer, but says too much.There is no need to identify the domain as the set ${0,1}$ nor the codomain as the real or complex numbers. – Ethan Bolker Jun 15 '20 at 01:00
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the codomain $\supset {a,b}$. Could be just about any set so long as it contains $a$ and $b$ which could be any type of object. No reason to assume they are numbers. No reason at all. – fleablood Jun 15 '20 at 02:09