# Two languages such that $L_1 \cup L_2 \leq_m\, L_1 \cap L_2$ and two (other?) such that $L_1 \cap L_2 ≤_m\, L_1 \cup L_2$?

Are there languages $$L_1$$, $$L_2$$ such that such that $$L_1 \cup L_2\leq_m\, L_1\cap L_2,$$ and two other languages such that $$L_1 \cap L_2 \leq_m\, L_1 \cup L_2?$$ And if so, what are they? How can i go about finding these?

How about taking $$L_1=L_2$$? Then we will have both $$L_1 \cup L_2\leq_m\, L_1\cap L_2$$ and $$L_1 \cap L_2 \leq_m\, L_1 \cup L_2$$.
If you want to have $$L_1\not= L_2$$, you just try $$L_2=L_1\cup\{w\}$$, where $$w$$ is a string not in $$L_1$$.
If you are determined to have infinity elements in both $$L_1\setminus L_2$$ and $$L_2\setminus L_1$$, try $$L_1=L\cup aL$$ and $$L_2=L\cup bL$$, where $$L$$ is an infinite language over $$\Sigma$$ and $$a,b$$ are two new symbols not in $$\Sigma$$.
• My bad, it should be $$L_1 \cup L_2\leq_m\, L_1\cap L_2,$$ not $$L_1 \cup L_2\leq_m\, L_2\cap L_2,$$ but i like your solution, any solution for the edited one? – Karlberg Jun 10 '19 at 16:50
• The same solution works. In fact, I did not notice it was $L_2\cap L_2$ when I wrote the answer. – John L. Jun 10 '19 at 16:51