# finite compliment and discrete topology question

• Apr 15th 2009, 10:24 PM
r2dee6
finite compliment and discrete topology question
Let the topological space X be \$\displaystyle \Re\$ with the finite complement topology, and let the topological space Y be \$\displaystyle \Re\$ with the discrete topology. Define a function

f: X---> Y by f(x) = \$\displaystyle \Re\$ - {x}. So, for example, If
U = \$\displaystyle \Re\$ - C, where C = {\$\displaystyle {x_1,......,x_n}\$}, then f(U) = C

(a) Is f continuous?Why or why not?

(b) What is \$\displaystyle f^-1\$?Is it continuous or not?

(c) Is f a homeomorphism? Why or why not?

• Apr 17th 2009, 12:54 PM
aliceinwonderland
Quote:

Originally Posted by r2dee6
Let the topological space X be \$\displaystyle \Re\$ with the finite complement topology, and let the topological space Y be \$\displaystyle \Re\$ with the discrete topology. Define a function

f: X---> Y by f(x) = \$\displaystyle \Re\$ - {x}. So, for example, If
U = \$\displaystyle \Re\$ - C, where C = {\$\displaystyle {x_1,......,x_n}\$}, then f(U) = C

Quote:

(a) Is f continuous?Why or why not?
Let A be an infinite union of singleton sets in the topological space Y. A is open in the topological space Y.
However, \$\displaystyle f^{-1}(A)\$ is not necessarily open in X.
Thus, f is not continuous.
Quote:

(b) What is \$\displaystyle f^-1\$?Is it continuous or not?
f^-1 is continuous since the domain of f^-1 is given by the discrete topology (verify this)
Quote:

(c) Is f a homeomorphism? Why or why not?
f is not a homeomorphism, because f is not continuous.