Answer with Step-by-step explanation:
We are given that a function is a continuous on R
f:R[tex]\rightarrow [/tex]R
We have to prove that if function is continuous ton R iff inverse image of closed set H is closed.
Let H be a closed set and function is continuous then R-H is a opens set
[tex]f^{-1}(R-H)=f^{-1}(R)-f^{-1}(H)=R-f^{-1}(H)[/tex]=Open set
When function is continuous then inverse image of open set is open
Hence, [tex]f^{-1}(H) [/tex]is a closed set
Conversely,
Let inverse image of closed set H is closed
If H is closed set then R-H is open set
[tex]f^{-1}(R-H)=f^{-1}(R)-f^{-1}(H)=R-f^{-1}(H)[/tex]
When inverse image of closed set is closed then R-inverse image of H is opens set
When inverse image of open set is open then the function is continuous.
Hence, function is continuous.
Hence proved.
