If S is countable and nonempty, prove their exist a surjection g: N --> S
1 answer:
Answer with Step-by-step explanation:
We are given S be any set which is countable and nonempty.
We have to prove that their exist a surjection g:N
Surjection: It is also called onto function .When cardinality of domain set is greater than or equal to cardinality of range set then the function is onto
Cardinality of natural numbers set =( Aleph naught)
There are two cases
1.S is finite nonempty set
2.S is countably infinite set
1.When S is finite set and nonempty set
Then cardinality of set S is any constant number which is less than the cardinality of set of natura number
Therefore, their exist a surjection from N to S.
2.When S is countably infinite set and cardinality with aleph naught
Then cardinality of set S is equal to cardinality of set of natural .Therefore, their exist a surjection from N to S.
Hence, proved
You might be interested in
Given:
Vertices of a square are A(-4,6), B(5,6) C(4,-2), and D(-5,-2).
To find:
The intersection of the diagonals of square ABCD.
Solution:
We know that diagonals of a square always bisect each other. It means intersection of the diagonals of square is the midpoint of diagonals.
In the square ABCD, AC and BD are two diagonals. So, intersection of the diagonals is the midpoint of both AC and BD.
We can find midpoint of either AC or BD because both will result the same.
Midpoint of A(-4,6) and C(4,-2) is
Therefore, the intersection of the diagonals of square ABCD is (0,2).