# Countably infinite composition of injective functions.

Just out of curiosity that came from a topology homework assignment where I had to show the composition of 3 injective functions was injective.

Suppose $f_i : A_i \mapsto A_{i+1}$ were injective where $i \in \mathbb{N}$. I know that the composition of n such functions, i.e. $\bigcirc_{i=1}^n f_i$, is injective. But what about $\bigcirc_{i=1}^\infty f_i$? Is this an injective function?

Also, is composition bounded to countability? That is, I think that the definition of composition limits us to an ordering which means that we can't have a composition of uncountably many functions. Is this true?

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 How do you define $\bigcirc_{i=1}^\infty f_i$? – joriki Sep 10 '12 at 16:12 $\left( \cdots \circ f_n \circ f_{n-1} \circ \cdots \circ f_1 \right)$ – torrho Sep 10 '12 at 16:17 Yes, I thought as much, but what does that mean? If $f_i:\mathbb R\to\mathbb R,x\to-x$, then what is $\bigcirc_{i=1}^\infty f_i$? – joriki Sep 10 '12 at 16:19 oh wow. Thats almost analogous to the "chicken or the egg" delimma... So then, perhaps it is not possible to define deterministic infinite composition, correct? – torrho Sep 10 '12 at 16:21 Well, it's not possible in that generality. For $f_i:X\to X$ with a topology defined on $X$, you could define $\bigcirc_{i=1}^\infty f_i=\lim_{n\to\infty}\bigcirc_{i=1}^n f_i$ whenever that limit exists. – joriki Sep 10 '12 at 16:26
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