A question about the cardinality of a set of functions with finite support where the domain of each function has cardinality aleph-null.

Question

Suppose that $M$ is a well ordered set with at least two elements and that $L$ is a well ordered set with the same cardinality as the natural numbers. Let $[L, s.f, M]$ be the set of all functions $f:L\to M$ with finite support. Is there an injection from $L$ to $[L, s.f,M]$? The support of $f$ is the set of all $x\in L$ such that $f(x)\neq a$, where $a≤y$, for all $y\in M$. My thought is that for each $n$ of $L$, there is a function $f*:L\to M$ such that $f*(x)=y$, for all $x≤n$ (where $y$ is not the least element of $M$) and for all $x>n,f*(x)=a$ (where $a$ is the least element of $M$). Since for each such $n$, the support of $f*$ would have the cardinality $n$, namely a finite cardinality, it would follow that there is a countably infinite number of these functions from $L$ to $M$, and so, an injection from $L$ to $[L, s.f, M]$. Am I missing something important?

Thanks.

Answer 1

First of all, since $M$ is just a linearly ordered set, there is no reason why $M$ needs to have a least element. So you definition of support is not well-defined if $M$ does not have a least element. But let me assume that $M$ does have a least element $a$. Since $M$ has at least two elements, let $b$ be any other element of $M$.

For each $x \in L$, let $f_x : L \rightarrow M$ be defined by $$f_x(u) = \begin{cases} a & \quad u \neq x \\ b & \quad u = x \end{cases}$$

It is clear that $f_x$ is a function with finite support. If $x \neq y$, then clealy $f_x \neq f_y$.

Define $\Phi$ by $\Phi(x) = f_x$. $\Phi$ is an injection of $L$ into the set of all functions from $L$ to $M$ with finite support.

Answer 2

The following was edited to take into account that $M$ is not necessarily assumed countable. I use the symbol $0$ to refer to the "null" element of $M$, whose inverse image is the complement of the "support".

Let $M^L_n = \left \{ f \in M^L : \text{supp}(f) \text{ has cardinality } n \right \}$. There is a bijection from $L^n \times M^n$ to $M^L_n$ which is given by

$$f(l_1,\dots,l_n,m_1,\dots,m_n)(l) = \begin{cases} m_i, & l = l_i \\ 0, & \text{otherwise} \end{cases}$$

Intuitively this bijection chooses $n$ elements of $L$ to be the support of $f$, then independently chooses elements of $M$ to be the image of those points.

Now provided the axiom of choice, we have that, for $n>0$, $|L^n \times M^n| = \max \left \{ |M^n|,|L^n| \right \} = \max \{ |M|,|L| \}$. Both of these equalities require that $L$ is assumed to be infinite.

Finally, the set $M^L_f = \left \{ f \in M^L : \text{supp}(f) \text{ is finite} \right \}$ is exactly $\bigcup_{n=0}^\infty M^L_n$. This union is disjoint. Therefore $M^L_f$ has cardinality $\max \{ \aleph_0,|M|,|L| \} = \max \{ |M|,|L| \}$, again because $L$ is assumed to be infinite (and because $\aleph_0$ is the smallest transfinite cardinal).