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mathboy
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[SOLVED] Find a one-to-one function
Let A1, A2, A3... be infinite pairwise disjoint sets. Find an injective function
f : A1 U A2 U A3 U... -> A1 x A2 x A3 x...
For all i, I choose fixed elements a_i in each A_i. If x is in A_n, I tried
f(x) = (a1, a2,...,a_n-1, x, a_n+1,...).
But this doesn't work because, e.g., f(a1)=f(a2). So I added the extra rule
f(a_n) = (a1, a2,...,a_n-1, b_n, a_n+1,...),
where a_n not= b_n. But this still doesn't work because f(a_n)=f(b_n). Can someone help me find an injective function that works?
Or at least prove that such an injective function exists using the axiom of choice (without simply stating that the cardinality of the domain is smaller than the cardinality of the range, because using this fact is not allowed in my question, unless I prove it using the axiom of choice).
Let A1, A2, A3... be infinite pairwise disjoint sets. Find an injective function
f : A1 U A2 U A3 U... -> A1 x A2 x A3 x...
For all i, I choose fixed elements a_i in each A_i. If x is in A_n, I tried
f(x) = (a1, a2,...,a_n-1, x, a_n+1,...).
But this doesn't work because, e.g., f(a1)=f(a2). So I added the extra rule
f(a_n) = (a1, a2,...,a_n-1, b_n, a_n+1,...),
where a_n not= b_n. But this still doesn't work because f(a_n)=f(b_n). Can someone help me find an injective function that works?
Or at least prove that such an injective function exists using the axiom of choice (without simply stating that the cardinality of the domain is smaller than the cardinality of the range, because using this fact is not allowed in my question, unless I prove it using the axiom of choice).
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