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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, is a union of open...