What does w ∈ a, b* mean?

Question

What does $w ∈ a, b^*$ mean?

The context is the language of an automation, which is $L=\{w∈ a, b^*|:|w|$ is even and the central symbols of $w$ are $aa\}$

I really don't understand what $w ∈ a, b^*$ mean, I think it means that $w$ can be $a$ or $b^*$ but that makes no sense because $w$ has to have the centraly symbols $aa$.

[EDIT] I tried to applly the pumping lemma to show that the language is not regular. I chose the string $w=b^paab^p$ and I divided it as $x=b^r$, $y=b^s$ and $z=b^{p-r-s}aab^p$. For $i=0$ I get $xy^iz=xz=b^{p-s}aab^p$ so $|b^{p-s}|<|b^p|$ because $|s|>0$ but this proves nothing because $p$ could be 4 and $s$ could be 2 so the word could still be even-lenght, how can I do?

Answer 1

It’s almost certainly missing a pair of curly braces and should be $$L=\big\{w\in\{a,b\}^*:|w|\text{ is even and the central symbols of }w\text{ are }aa\big\}\;.$$

Here $\{a,b\}^*$ is the set of all finite strings of symbols from the set $\{a,b\}$, i.e., the set of all finite strings composed exclusively of the letters $a$ and $b$, including the empty string. $L$ is the set of all such strings with even length and $aa$ as the middle two characters, so $L$ includes the strings $aa$, $aaaa$, $aaab$, $baaa$, $baab$, $aaaaaa$, $abaabb$, etc.

Added: For showing that $L$ is not regular using the pumping lemma, you were on the right track when you took $w=b^paab^p$, where $p$ is the pumping length. If $L$ were regular, you could write $w$ as $xyz$, where $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for each $k\ge 0$. Since $|xy|\le p$, $xy$ must, as you saw, be a string of $b$s, say $x=b^r$ and $y=b^s$, where $r+s\le p$ and $s\ge 1$, so for any $k\ge 0$ we have $$xy^kz=b^{p+(k-1)s}aab^p\;.$$

Now take $k=2$: $xy^2z=b^{p+s}aab^p$. And $p+s>p$, so the word cannot have $aa$ at the centre: there is only one subword $aa$, and there are more characters before it than after it. (It’s true that if $s$ is odd, you can eliminate this word even more easily, since its length is then odd, but showing that the word doesn’t a central segment $aa$ does the job no matter what $s$ is.)

Answer 2

Your pumping lemma proof if fine. Let's write it out in detail:

By contradiction. Assume $L = \{w \in \{a, b\}^* \colon \text{middle symbols are $aa$}\}$ is regular. Then $L$ satisfies the pumping lemma for regular languages. Call the constant of the lemma $N$, select $\sigma = b^N a a b^N$, so that $\lvert \sigma \rvert = 2 N + 2 \ge N$, so we can write $\sigma = \alpha \beta \gamma$ with $\lvert \alpha \beta \rvert \le N$ and $\beta \ne \varepsilon$ such that for all $k \ge 0$ it is $\alpha \beta^k \gamma \in L$. But $\alpha \beta$ are just $b$s, repeating $\beta$ thrice gives a string that has $b$ in the middle, contradiction.