Consider the languages L1 = Φ and L2 = {a}. Which one of the following…

Concept

Three operations on languages govern this problem. (1) Concatenation: for languages X and Y, XY = { xy : x ∈ X, y ∈ Y }; if either factor is the empty set, no pair (x, y) exists, so X·∅ = ∅·Y = ∅. (2) Kleene star: X* = X0 ∪ X1 ∪ X2 ∪ … , and X0 = {ε} by definition (the empty concatenation), so {ε} is ALWAYS a member of any starred language — even ∅* = {ε}. (3) Union just collects the members of both sides.

Application

1. Evaluate the concatenation L1L2 = Φ·{a}. Since L1 = Φ has no strings, there is no pair to concatenate, so Φ·{a} = Φ.

2. Evaluate the star L1* = Φ*. The 0-length concatenation contributes ε, so Φ* = {ε}, regardless of Φ being empty.

3. Take the union: L1L2 ∪ L1* = Φ ∪ {ε} = {ε}.

Cross-check

The expression contains the symbol a only inside L2, which is wiped out the moment it is concatenated with the empty language; nothing else can introduce a, so the result cannot contain any string with a. The star term can only contribute ε. Hence the final language is exactly {ε} — a single-element set containing the empty string, not the empty set and not {ε, a}.