if A is decidable then B is decidable too Assume that a language A is reducible to language B. The claim is true?

if A is decidable then B is decidable too.

The correct answer is:


  This claim is wrong. If A is e.g. the empty language (which is clearly decidable) and B is $A_{TM}$, then surely ∅ is reducible to $A_{TM}$, but $A_{TM}$ is undecidable. The claim is true the other way round: If B is decidable then A is decidable too.


I have problem to understand two points:


A is e.g. the empty language (which is clearly decidable)


Why empty language is ("clearly") decidable and $L_∅$ is not?, where,  $$L_\emptyset = \{\langle M\rangle \mid M \text{ is a Turing Machine and }L(M)=\emptyset\}.$$


**A is reducible to $A_{TM}$ **


I have difficulties to image a reduction from $∅$ to $A_{TM}$, where $A_{TM} = \{&lt;M,w&gt; | \ M \ is \ a \ TM \ and \ M \ accepts \ w\}$
By definition, a language $L$ is decidable if there exists a TM $M\mid L = L(M)$ deciding it. Consider a TM that rejects on all inputs (for example, one where $q_o = q_{\text{rej}}$). The language of this TM is $\emptyset$, so $L=\emptyset$ is decidable.
With reductions, in general if $A\leq B$ ($A$ reduces to $B$), then $B$ is at least as hard as $A$ with respect to decidability. An undecidable language $A_{TM}$, regardless of what it actually is, is harder to decide than the decidable language $\emptyset$ by definition. The specifics of the reduction don't matter; the only information we needed was given in the problem statement (that $A$ is reducible to $B$).