Given an alphabet Σ the Kleene closure of Σ is a language given by

Σ^{*} = Σ^{0} ∪ Σ^{1} ∪ Σ^{2}……….

**For example,**

1. if Σ = {X}, then

Σ* = {**ε**, x, xx, …..}

2. if Σ = {0, 1}, then

Σ^{*} = {ε,0, 1, 00, 01, 10, 11, ……}

If L is language, then L^{*} is the set of all finite strings formed by concatenating Words from L. Any word can be used any number of times including zero time. **ε** is a member of L^{*}.

**Example : **If L = {aa,b} then

In this example, we should assume ‘aa’ is first character and ‘b’ is second character.

L^{*} = {ε, b, aa, bb, bbb, baa, aab, aabb, bbaa, ……}

L^{*} always contains an epsilon.

It may be noted that a always appears in pair.

If Kleene closure Σ = **Ф** (an empty language)

then Σ^{*} = {ε}

L^{+} is known as the positive closure of L. L^{+} is the set of all finite strings formed by concatenation words from L. Any word can be used one or more times.

**Example :** If L = {aa, b} then

L^{+} = {b, aa, bb, bbb, baa, aab, ……}

L^{+} always not contains an epsilon.

**Question 1.**

#### Let L be a language. It is clear from the definition that L^{+ }⊆ L*. Under what circumstances are they equal ?

**Solution :**

L^{*} always contains an epsilon.

L^{+} always not contains an epsilon.

Thus, the only difference between L^{+ }and L^{*} is that L^{+} may or may not contain an **epsilon **but L^{*} will always contain an **epsilon**. **(epsilon mean empty )**

If L itself contains an epsilon the there will be no difference between L^{+ }and L^{*} .

**Recommended:**

- Automata Theory and Formal Languages
- Recursive Definition of a Language
- Finite Representation of Language