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text: '[summary: Here we give three proofs that the only [3jz set] which satisfies the [-5zr] is the [-5zc] itself.]\n\nHere, we will prove that the only [3jz set] which satisfies the [-5zr] is the [-5zc] itself.\nThis will tell us that defining the empty set by this [-600] is actually a coherent thing to do, because it's not ambiguous as a definition.\n\nThere are three ways to prove this fact: one way looks at the objects themselves, one way takes a more maps-oriented approach, and one way is sort of a mixture of the two.\nAll of the proofs are enlightening in different ways.\n\nRecall first that the universal property of the empty set is as follows:\n\n> The empty set is the unique set $X$ such that for every set $A$, there is a unique function from $X$ to $A$.\n(To bring this property in line with our usual definition, we denote that unique set $X$ by the symbol $\\emptyset$.)\n\n# The "objects" way\n\nSuppose we have a set $X$ which is not empty.\nThen it has an element, $x$ say.\nNow, consider maps from $X$ to $\\{ 1, 2 \\}$.\n\nWe will show that there cannot be a unique [-3jy] from $X$ to $\\{ 1, 2 \\}$.\nIndeed, suppose $f: X \\to \\{ 1, 2 \\}$.\nThen $f(x) = 1$ or $f(x) = 2$.\nBut we can now define a new function $g: X \\to \\{1,2\\}$ which is given by setting $g(x)$ to be the *other* one of $1$ or $2$ to $f(x)$, and by letting $g(y) = f(y)$ for all $y \\not = x$.\n\nThis shows that the universal property of the empty set fails for $X$: we have shown that there is no unique function from $X$ to the specific set $\\{1,2\\}$.\n\n# The "maps" ways\n\nWe'll approach this in a slightly sneaky way: we will show that if two sets have the universal property, then there is a [499 bijection] between them. %%note: The most useful way to think of "bijection" in this context is "function with an inverse".%%\nOnce we have this fact, we're instantly done: the only set which bijects with $\\emptyset$ is $\\emptyset$ itself.\n\nSuppose we have two sets, $\\emptyset$ and $X$, both of which have the universal property of the empty set.\nThen, in particular (using the UP of $\\emptyset$) there is a unique map $f: \\emptyset \\to X$, and (using the UP of $X$) there is a unique map $g: X \\to \\emptyset$.\nAlso there is a unique map $\\mathrm{id}: \\emptyset \\to \\emptyset$. %%note: We use "id" for "identity", because as well as being the empty function, it happens to be the identity on $\\emptyset$.%%\n\nThe maps $f$ and $g$ are inverse to each other. Indeed, if we do $f$ and then $g$, we obtain a map from $\\emptyset$ (being the domain of $f$) to $\\emptyset$ (being the image of $g$); but we know there's a *unique* map $\\emptyset \\to \\emptyset$, so we must have the composition $g \\circ f$ being equal to $\\mathrm{id}$.\n\nWe've checked half of "$f$ and $g$ are inverse"; we still need to check that $f \\circ g$ is equal to the identity on $X$.\nThis follows by identical reasoning: there is a *unique* map $\\mathrm{id}_X : X \\to X$ by the fact that $X$ satisfies the universal property %%note: And we know that this map is the identity, because there's always an identity function from any set $Y$ to itself.%%, but $f \\circ g$ is a map from $X$ to $X$, so it must be $\\mathrm{id}_X$.\n\nSo $f$ and $g$ are bijections from $\\emptyset \\to X$ and $X \\to \\emptyset$ respectively.\n\n# The mixture\n\nThis time, let us suppose $X$ is a set which satisfies the universal property of the empty set.\nThen, in particular, there is a (unique) map $f: X \\to \\emptyset$.\n\nIf we pick any element $x \\in X$, what is $f(x)$?\nIt has to be a member of the empty set $\\emptyset$, because that's the codomain of $f$.\nBut there aren't any members of the empty set!\n\nSo there is no such $f$ after all, and so $X$ can't actually satisfy the universal property after all: we have found a set $Y = \\emptyset$ for which there is no map (and hence certainly no *unique* map) from $X$ to $Y$.\n\nThis method was a bit of a mixture of the two ways: it shows that a certain map can't exist if we specify a certain object.',
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