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**Moo** And the union is not necessarily in bijection with N. We know it's countable, but we don't know if it has the same cardinal as N (bijection)

You can automatically conclude that E is countable with injective functions, which is why you don't have to bother with bijections (I find injections easier to deal with than bijections...)

Moreover, E_n is countable means that there exists an injection from E_n to N, not that there exists a bijection.

It's more general, since a bijection is an injection.

It's a messy explanation, but I hope it helps you...