What does it mean to map a function from the null set to any set $X$, i.e., $\emptyset \rightarrow X$? I'm reading Terry Tao's book on Analysis 1.
I'm having trouble understanding how a function can be mapped as $\emptyset \rightarrow X$.
After stating the following map he said that all such maps for each set $X$ are equal, which I did not seem to understand.
Any help clearing the concept will be appreciated. 
 A: A map $f\colon A\to B$ is determined by its graph, which is the set subset of $A\times B$ consisting of the pairs $(a,f(a))$ for all $a\in A$. Indeed a subset $R$ of $A\times B$ is the graph of a map $A\to B$ if and only if for each $a\in A$ there is exactly one $b\in B$ such that $(a,b)\in R$. The corresponding map then has $f(a)=b$.
Now when $A=\varnothing$, the cartesian product $\varnothing\times B$ is empty, so the only possible graph of a function would be empty as well. And indeed this does satisfy all the properties: for each $a\in \varnothing$ there is a unique $b\in B$ such that $(a,b)\in\varnothing\times B = \varnothing$. This is satisfied since there is no $a\in\varnothing$ that could serve as a counter example.
We conclude that for any set $B$ there is only one map $\varnothing\to B$ and its graph is the empty set.
A: In general (or at least in set-theory) a function $f:A\to X$ is a subset of $A\times X$ such that for every $a\in A$ there is a unique $x\in X$ such that $(a,x)\in f$.
Now observe that for every set $X$ we have:$$\varnothing\times X=\varnothing$$
Then observe that $\varnothing$ itself is the only subset of this set.
Further if you pose the question: "is this subset a function?" then the answer on it is clearly: "yes".
Shown is now that for every set $X$ we have exactly one function $\varnothing\to X$ and secondly in all cases this function is $\varnothing$ declaring what Terry means if he says that all such maps are equal.
A: A mapping $f:A\rightarrow B$ is a relation $f\subseteq A\times B$, which is (1) left-total, i.e., for each $x\in A$ there exists a $y\in B$ such that $(x,y)\in f$, and (2) right-unique, i.e., if $a\in A$ and $b,b'\in B$ with $(a,b),(a,b')\in f$, then $b=b'$.
Thus for each $a\in A$ there is exactly one $b\in B$ with $(a,b)\in f$, written $f(a)=b$.
The ''empty'' mapping $f:\emptyset\rightarrow X$ exists by logical reasons:
(1) Left-total: $\forall a: a\in A\Rightarrow \exists b: [b\in B\wedge (a,b)\in f].$ 
This implication holds since the premise ''$a\in \emptyset$'' is false and so the implication is true.
Similar for right-uniqueness.
A: Formally, a function $f \colon M \rightarrow N$ is defined as a subset of $M \times N$, namely the graph. The function $\emptyset \rightarrow X$ is exactly the function that corresponds to the empty graph, which is the only possible graph when the domain is empty. Therefore this function is probably more easily understood on formal terms than the more intuitive notion $\emptyset \rightarrow X$.
