I have trouble understanding this statement in my lecture notes on Cardinality.
For all sets $X$ and all $n \in \mathbb{N}$, $Maps([n+1], X) \approx Maps([n], X) \times X$.
$[n]$ is defined to be the set $\{0,1,2,3, \dots n-1\}$
I have trouble understanding this statement in my lecture notes on Cardinality.
For all sets $X$ and all $n \in \mathbb{N}$, $Maps([n+1], X) \approx Maps([n], X) \times X$.
$[n]$ is defined to be the set $\{0,1,2,3, \dots n-1\}$
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the map $$Map([n+1],X)\to Maps([n], X) \times X, f\to (f/[n], f(n+1))$$ Where $f/[n] $ denotes the restriction of $f$ to the subset $][n] $ of $[n+1]$ is bijective. That is a function is uniquely determined, once we know its restrictions to a subset and its complement, and given any two complementary subsets,as domains of two maps, we can extend the two maps in a unique way to a map defined on their union.