# Construct a decidable set $B$ such that $B \neq A_w$ for any $w \in \Sigma^\star$

I've been stuck on this problem for a while. Any hints would be appreciated!

Let $$A \subseteq \Sigma^\star$$ be decidable. Given $$w \in \Sigma^\star$$, define $$A_w = \{x \in \Sigma^\star\:|\: \langle x, w \rangle \in A\}.$$ Construct a decidable set $$B$$ such that $$B \neq A_w$$ for any $$w \in \Sigma^\star$$.

• What is "$\langle x, w \rangle$"? is it the concatenation of $x$ and $w$? – John L. Apr 10 '19 at 5:10
• It is the encoding of the tuple $(x, w)$. You can think of it as $x \# w$, where $\#$ is a symbol in $\Sigma$. – K.rar Apr 10 '19 at 5:12
• I'm sorry, I wasn't aware of that. Thank you for letting me know and for your answer! – K.rar Apr 10 '19 at 6:10

Here is a hint. Let $$w\in B$$ if $$w\not\in A_w$$ and let $$w\not\in B$$ if $$w\in A_w$$.