I know that to show this I have to show there’s an injection from the $\varnothing$ to $\{\varnothing\}$. But how can one even define a function on the $\varnothing$?
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1What's the definition of a function? – Matthew Towers Nov 17 '19 at 16:30
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A mapping from one set to another – Partey5 Nov 17 '19 at 16:32
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2No, the formal definition. – Matthew Towers Nov 17 '19 at 16:32
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Of course, it is the empty function; $;\varnothing\times{\varnothing}$ is the empty set. – Bernard Nov 17 '19 at 16:33
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1Recall the definition of "function from $X$ to $Y$": A set $f\subset X\times Y$ such that for every $x\in C$ there exists exactly one $y\in Y$ with $(x,y)\in f$. It follows that if $A$ is any set then $\emptyset$ is an injective function from $\emptyset$ to $A$. – David C. Ullrich Nov 17 '19 at 16:33
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Formally, a function f from a set X to a set Y is defined by a set G of ordered pairs (x, y) such that x ∈ X, y ∈ Y, and every element of X is the first component of exactly one ordered pair in G. – Partey5 Nov 17 '19 at 16:34
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But if $X$ = emptyset then there is no $x \in X$ – Partey5 Nov 17 '19 at 16:38
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In the past I thought there exists a unique $f:A \rightarrow {1}$ defined by $a \rightarrow 1$ for all $a \in A$ where $A$ is non-empty set. Does the emptyset function mean I am wrong and that $f$ is not unique. – Partey5 Nov 17 '19 at 16:44
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1No, you're right. A function on A has to contain an ordered pair with first element a for each a in A, so there's no "empty function" on a non empty set. But there is an empty function on the empty set. – Matthew Towers Nov 17 '19 at 16:59
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"But if X = emptyset then there is no x∈X" EXACTLY! So for any $b \in Y$ there is no $(x,b) \in \emptyset \times Y$. And if there are $0$ such $(x,b)$ then there can not be more than one! So $\emptyset: \emptyset \to Y$ is injective because for every $y \in Y$ there is at most one (in fact there are zero) $(x,y) \in \emptyset \subset \emptyset \times Y$. – fleablood Nov 17 '19 at 16:59
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"In the past I thought there exists a unique f:A→1 defined by a→1 for all a∈A where A is non-empty set. " And that is true FOR ALL $a\in \emptyset$. That is trivially and vacuously true for all $a \in \emptyset$ because there aren't ANY $a\in \emptyset$. – fleablood Nov 17 '19 at 17:02
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A function, $f:X\to Y$ is a subset of a cartesian product, $X\times Y$ where each element $a \in X$ is "represented" precisely once.
$h:\emptyset \to \{\emptyset\} \subseteq \emptyset \times \{\emptyset\} = \{(x,y)|x \in \emptyset, y \in \{\emptyset\}\}$. But as $\emptyset$ is empty $x \in \emptyset$ is impossible so so $\emptyset \times \{\emptyset\} = \{(x,y)|x \in \emptyset, y \in \{\emptyset\}\} = \emptyset$.
So $h= \emptyset$. Is $\emptyset$ injective? What does that mean?
$f\subseteq X\times Y$ is injective if for any $b\in Y$ there is at most one $(x,b)\in f$.
So for any $b \in \{0\}$ how many $(x, b) \in h = \emptyset$. Well, zero as $h$ is empty. So is there at most one? Well, there sure as heck aren't more than one!
....
The emptyset is always an empty function from $\emptyset \to $ any set $X$ and it is vacuously injective.
Vacuous statements are, admittedly, irritating and confusing and you have my empathy. If it helps, you can put a mental close in most definitions as "An $x$ is GLOOP if when it exists GOBBLETY occurs. And when it doesn't exist it is GLOOP by default. That's kind of cheating but it helps.
The old stumbling block is "Every element in the emptyset is red". That is true. Every element of the emptyset, all zero of them, are red because there aren't any elements. But if we say "Every element in the emptyset, if it exists, is red" that's very clear. Unfortunately it's inaccurate because the element doesn't exist.... well, it's still technically accurate to say it is red.... sigh... it gets easier.
Althernatively, Every element is GLOOP $\iff $ there is no element that is not GLOOP. It's very clear that the RHS is always true for an emptyset as there no elements at all.
fleablood
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What’s confusing me is my definitions of injectivity have always involved elements in the domain... of which there aren’t any now! – Partey5 Nov 17 '19 at 17:05
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Right.... but that makes all (postive) statements vacuously true. "If $a \in \emptyset$ then $a$ is GLOOP." That must be true because 1) a false hypothesis ($a\in \emptyset $ is always false) leads to any conclusion, and 2) there is not counterexample: There are no $a\in \emptyset$ where $a$ isn't GLOOP. ... admittedly empty functions are hard to imaging but... well think of it like this... there aren't any elements of $\emptyset$ that aren't* being mapped to $Y$ so all elements of $\emptyset$ are being mapped to $Y$... – fleablood Nov 17 '19 at 17:13
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"Why have you written h to be a proper subset in line 3?" Notation varies by from person to person and text to text but, I, me, fleablood, do not consider $\subset$ to mean proper subset. If I wish to indicate proper subset, I would write $\subsetneq$. However, if I were writing in pen, if a wanted subset possibly not proper, I'd write $\subseteq$. But when I write in mathjax, I think "subset" so I write $\subset$ which comes out $\subset$. Sorry if that is confusing. – fleablood Nov 17 '19 at 17:18
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That’s ok. One more question. Is the function from the empty set to the emptyset then injective also by your reasoning – Partey5 Nov 17 '19 at 17:23
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Yes..... $h: \emptyset \to Y = \emptyset$ is an injective (empty) function for all $Y$. Even if $Y = \emptyset$. Worth noting that $h: \emptyset \to Y$ is never surjective (obviously) unless $Y$ is empty. Then $h: \emptyset \to \emptyset $ is bijective (vacuously). – fleablood Nov 17 '19 at 17:38
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@Anteater23. I've tried an explanation below, with an emphasis on the logical aspect of the problem. – Nov 17 '19 at 20:44
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General idea: in order 2 conditions to be both satisfied, it is already necessary the first one to be satisfied.
So if non-injectivity requires 2 conditions and if the first one cannot be satisfied, non-injectivity does not hold , and therefore, automatically , injectivity holds.
Let f be a function from { } to { { } }.
Let me " define" f arbitrarily like this : for all x, f(x) = 2x.
Now, whatever its " definition" may be, f is a relation, and therefore, in this case, a subset of the cartesian product :
"{ } cross { { } }"
that is , the set of all pairs (x,y) such that (1) x belongs to Empty Set and (2) y belongs to { { } }.
But, since condition (1) cannot be satisfied, a fortiori, there is no pair that satisfies both conditions, so my cartesian product is the Empty set.
I know that function f, that is a relation, is a subset of a cartesian product that is empty. But the only subset of the empty set is the empty set itself. So : function f such that f(x)=2x is identical to the Empty set.
( That would have been true whatever formula I would have used to "define" f).
Now, is f injective?
I know that f is injective just in case :
IF a pair (a, c) belongs to f ( whatever this pair may be) , THEN there is no pair ( b, c) belonging to f with element a different from element b.
( indeed, if we had at the same time (a, c) and ( b,c) with a different from b, then 2 diffetent elements of the domain would have the same image, and f would not be injective).
You know that a statement of the form (P --> Q) is false just in case P is true and Q is false. Let's apply this logical fact to our definition of injectivity which is an IF...THEN statement.
Let's see whether it is possible for f not to be injective.
For f NOT to be injective , it would be required that :
(1) there is actually a pair (a, b) belonging to f [ this is the " P is true" part]
AND
(2) there is ( at least) a pair (b, c) belonging to f with a different from b [ this is the "Q is false" part, for, remember, Q was " there is no pair..." ]
But it is impossible that both consitions hold, since it is already impossible the first one to hold: f being empty, there is no pair (a,c) that belongs to f, since there is nothing that belongs to f.
So the proposition " f is NOT injective" is FALSE ( and even necessarily false).
Consequently, the proposition " f is injective" must be true.