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Robert · Accepted Answer

Question1
Solution: 
This is right property 
The necessity of property for a continuous function is obvious. Suppose 
then that f possesses these property and let V Y be an open set. 
 Let be such that . Then 
showing that is a union of open sets and therefore open.
Or in other way this can be proved as 
As stated int(A) is open in Y, 
So that )]([1 AIntf   is open in X, 
Since )()]([ 11 AfAIntf    and )]([ 1 AfInt   is the greatest open set containing )(1 Af   
Thus 
))(())(( 11 AfIntAIntf  
This is wrong property 
let f be a constant function, say with value 0,from the reals to itself. Then f is continuous, 
but for A={0} we have Int(A)=∅ and so is its inverse image.
But )(1 Af  =R which has itself as the interior. So would say R⊆∅ as per property, which 
is nonsense 
Hence the property is wrong.
Counter part 
Suppose that f is continuous and let A   Y be a closed set. Then 
)()( 11 AYfXAf    is also closed since Y-A is open and continuity of f forces 
)(1 AYf   to be open also. 
Conversely, suppose that f has the property such that )(1 Bf  is a closed subset of X for 
any closed subset B of Y and let VY be any open set. 
Then )()( 11 VYfXVf   is open since Y-V and )(1 VYf  are both closed 
Let B be any subset of Y. Since B B , we obtain )()( 11 AfAf   . By the property of f, 
the set )(1 Af  is closed but bar of )(1 Af  since the set is the smallest closed set 
containing )(1 Af  , the inclusion bar of )(1 Af  )(1 Af  is immediate.
Solution: By the property 1 proved earlier,

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