There are two disjoint sets `S_1` and `S_2` where
`S_1={f(1),f(2),f(3)…..........}`
`S_2={g(1),g(2),g(3)….........}` such that `S_1 uu S_2` forms the set of natural numbers.
Also `f(1) lt f(2) lt f(3)…........`& g(a)` lt g(2) lt g(3)` and f(n)=g(g(n))+1 then what is g (1) ?

Options

A. 0
B. 1
C. 2
D. can't be determined

Answer

To solve the problem step by step, we need to analyze the given information about the sets \( S_1 \) and \( S_2 \), the functions \( f(n) \) and \( g(n) \), and their relationships.

### Step 1: Understand the Sets and Functions
We have two disjoint sets:
- \( S_1 = \{ f(1), f(2), f(3), \ldots \} \)
- \( S_2 = \{ g(1), g(2), g(3), \ldots \} \)

It is given that \( S_1 \cup S_2 \) forms the set of natural numbers \( \mathbb{N} \). This means every natural number is either in \( S_1 \) or \( S_2 \), but not in both.

### Step 2: Analyze the Properties of the Functions
We know:
- \( f(1) < f(2) < f(3) < \ldots \) (increasing)
- \( g(1) < g(2) < g(3) < \ldots \) (increasing)
- The relationship \( f(n) = g(g(n)) + 1 \)

### Step 3: Establish Relationships
From the equation \( f(n) = g(g(n)) + 1 \), we can deduce that:
- \( f(n) > g(g(n)) \) because \( g(g(n)) + 1 \) is always greater than \( g(g(n)) \).

### Step 4: Compare \( f(n) \) and \( g(n) \)
Since \( f(n) > g(g(n)) \) and \( g(n) \) is increasing, we can conclude:
- \( f(n) > g(n) \) for all \( n \).

### Step 5: Determine the Smallest Element
Since \( S_1 \) and \( S_2 \) are disjoint and together they cover all natural numbers, the smallest element in \( S_1 \cup S_2 \) must be 1 (the smallest natural number).

### Step 6: Identify \( g(1) \)
Since \( g(1) \) is the smallest element in \( S_2 \) and \( S_2 \) must contain natural numbers that are not in \( S_1 \), we can conclude:
- The smallest number in \( S_2 \) (which is \( g(1) \)) must be 1, because if \( g(1) \) were 0 or any number less than 1, it would not be a natural number.

### Conclusion
Thus, we conclude that:
\[ g(1) = 1 \]

### Final Answer
The value of \( g(1) \) is **1**.