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I just need some clarifications in the solution to the problem my teacher gave me below:

Suppose A and B are disjoint denumerable sets. Show that $A\cup B$ is denumerable by constructing a bijection.

proof: Since A and B are denumerable, then f:N->A and g:N->B are bijective.

h:N->$A\cup B$ is defined by :

h(n)= { $f(\frac{n+1}2)$ if n is odd and $g(\frac{n}2)$ if n is even)

From what I know, to now show bijection, I would have to prove surjectivity and injectivity. But my question here is why h(n) was defined like this. Why is my teacher saying h(n)= { $f(\frac{n+1}2)$ if n is odd and $g(\frac{n}2)$ if n is even). I was just wondering if the g and f could be switched. Like h(n)= { $g(\frac{n+1}2)$ if n is odd and $f(\frac{n}2)$ if n is even).

Jr194
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    Have you seen a zipper? They are taking one element from the sequence $f$, then one from $g$, then one from $f$ and one from $g$, ... – conditionalMethod Dec 08 '19 at 12:06
  • I kind of get what you mean. Is it because since we taking from f first, and g second that my teacher structured it this way? – Jr194 Dec 08 '19 at 12:08
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    You're right that it wouldn't matter if $f$ and $g$ were switched! The point is that we just need to alternate between picking elements from $A$ and $B$ in a zig-zaggy (or zippery) way, so that we are guaranteed to eventually hit all elements in both exactly once. It doesn't matter if we start with an element from $A$ or an element from $B$. – Jack Crawford Dec 08 '19 at 12:44

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