$f: \Sigma^* \mapsto \Sigma^*$ is a mapping reduction from language A to language B if $f$ is computable and for every $x \in \Sigma^*$, $x \in A \iff f(x) \in B$.

Take for instance $A = A_{TM}$ and $B = \Sigma^*$. Clearly A in contained in B but the identity function is not a mapping reduction since for every $x \not\in A_{TM}$, $f(x) \in \Sigma^*$. ]]>

in this question we were asked to prove or disprove that if A is contained in B then you can do a reduction from A to B.

The answer is that it's incorrect.

But why isn't that answer contradict the definition of reduction? the definition is that if you have a computable function from A to B for every member of A, you have a reduction.

clearly here there is a function like that - f(x) = x, and since A is contained in B then it satisfies the definitions… no?

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