Question

If `f`: R → {0, 1} is a continuous surjection map then `f^(-1) (0) ∩ f^(-1) (1)` is:

Options

• `phi`

• R

• {0, 1}

• None of these

Answer 1

`phi`

Explanation:

Since `f: R -> {0, 1}`

∴ Domain = R and Range = {0, 1}

Since each element of domains set have unique image either 0 or 1, therefore preimage of 0 and 1 have different element

∴ `f^(-1) (0) ∩ f^(-1) (1) = phi`