Respuesta :

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.

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