Is the union of finite and countably infinite sequence over alphabet $\Sigma=\{1\}$, countably infinite as well?

I understand this is similar a question to the one of finite and countably infinite strings over $\{0,1\}$, however I think that the previous argument is not valid in this case.

Now, countably infinite strings over $\{0,1\}$ can be of any kind, e.g.

$$e_i \in \Sigma^*, e_i=100000...$$ $$e_k \in \Sigma^*, e_k=110000...$$

However, I noticed that if I pose the same question with an alphabet of a character only, say $\{1\}$, there is only 1 infinite sequence that I can generate $11.....$, hence:

$$S = \Sigma^* \cup \{111..\}$$

From this assumption, I could map $0$ to the infinite $11...$ and assume that this set S is uncountable. Is this a valid argument? Can someone help me formalize this?


You haven't defined what you mean by "finite or countably infinite sequence" over an alphabet $\Sigma$, so let me assume that a "finite sequence" is a word over $\Sigma$, while a "countably infinite sequence" is an $\omega$-word over $\Sigma$, that is, a mapping from $\omega$ (the set of natural numbers) to $\Sigma$.

In your case, there is exactly one word of each length $0,1,2,\ldots,\omega$, so the set of "finite and countably infinite sequences" is countable, since the set $\omega + 1 = \omega \cup \{\omega\}$ of possible lengths (which is in one-to-one correspondence with the possible "finite and countable infinite sequences") is countable.

If you want to be very formal, you can show that $\omega + 1$ is countable by giving an explicit bijection between $\omega$ and $\omega + 1$, such as $$ 0 \mapsto \omega, \quad n+1 \mapsto n \text{ for all } n \geq 0. $$

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