Why is $\emptyset L = L\emptyset = \emptyset$ correct?

Question

I am taking a course on Automaton where I faced the algebraic laws of regular expressions.

First two were ok:

1. $\emptyset + L = L + \emptyset = L$: Union of a language $L$ with empty language gives the language $L$ itself.

2. $\epsilon L = L\epsilon = L$: Concatenation of empty string with strings belonging to certain language $L$ gives back the set of string of language $L$

But the third one is confusing me:

3. $\emptyset L = L\emptyset = \emptyset$

I felt it should be $\emptyset L = L\emptyset = L$, since concatenation of strings of language L with string of empty language which is essentially no strings at all (which essentially means nothing to concatenate), should give back the language $L$ itself. Where I am thinking wrong?

Answer 1

$L_1L_2$ means the set of all strings you can make by choosing one from $L_1$ and one from $L_2$ and concatenating the two strings. So how many ways can you choose a string from the empty set of strings and one from $L$? None at all! So there are no such concatenations and $\emptyset L = \emptyset$.

The key point is that $\emptyset$ is the language that contains no strings at all. The empty string is a string (it's the string with no letters at all) so the empty string is not a member of the empty language: $\{\epsilon\}\neq\emptyset$.

Answer 2

Just to complete David's nice answer with a simple observation:

Since $$ L_1L_2 =\{ uv \mid u\in L_1, v\in L_2\},$$

then, the size of this new language admits $$ |L_1L_2| \le |L_1|\cdot|L_2|$$ because, for each word in $L_1$ we add $|L_2|$ words to $L_1L_2$. Maybe some of the new words were already added before, which causes the inequality.

Then, it is immediate that $|\emptyset L| \le 0\cdot |L| \le 0$, thus $\emptyset L$ must be the empty language.