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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 $w \in \Sigma^\star$.

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

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