Question

(1) Let X = {0}U[2,3], and give X the topology Tx = {0,{0}, [2, 3], X}. (a) (10 points) Is X To? Briefly justify your answer. (b) (10 points) Is X Hausdorff? Briefly justify your answer. (c) (10 points) Is X Tz? Briefly justify your answer. (d) (10 points) Is D = ({0} Ⓡ {0}) U ([2, 3] x [2, 3]) a closed subset of X x X with the product topology? Briefly justify your answer.

Answer 1

Solution :

(1)

(a)
Consider the points 2 and 3 in X. If the space X is T0, then there exists an open set U which contains one of the two points but not the other.
However, the only nonempty open sets in the space X are {0} , [2,3] and X. {0} contains neither 2 nor 3, [2,3] contains both 2 and 3 and similarly, X contains both 2 and 3.
Thus, there is no such open set which contains exactly one of the two points 2 and 3.
Hence, X is not T0.

(b)
Since, any Hausdorff space is T0 and since the space X is not T0, hence X is not Hausdorff.

(c)
Again, since any T3 space is T0 and since the space X is not T0, hence X is not T3.

(d)
Consider X x X with the product topology.
A basis for the product topology on X x X is { U x V : U,V are open in X }

Claim : The complement of ({0} x {0}) U ([2,3] X [2,3]) in X x X is ({0} x [2,3]) U ([2,3] x {0}).
Proof of the claim : Let (x,y) $\in$ ({0} x [2,3]) U ([2,3] x {0}).
Then, (x,y) $\in$ {0} x [2,3] or (x,y) $\in$ [2,3] or {0}.
Thus, either x = 0 or y = 0 but not both.
Hence, (x,y) is not contained in ({0} x {0}) U ([2,3] X [2,3]).
Therefore, (x,y) $\in$ (X x X) \ ( ({0} x {0}) U ([2,3] X [2,3]) ).
Hence, ({0} x [2,3]) U ([2,3] x {0}) $\subseteq$ (X x X) \ ( ({0} x {0}) U ([2,3] X [2,3]) ).

Now, suppose that (x,y) $\in$ (X x X) \ ( ({0} x {0}) U ([2,3] X [2,3]) )
Then, (x,y) is not contained in ({0} x {0}) U ([2,3] X [2,3]).
Thus, {x,y} is neither a subset of {0} nor a subset of [2,3].
Since x,y are in X = {0} U [2,3], hence either (x $\in$ {0} and y $\in$ [2,3]) or (x $\in$ [2,3] and y $\in$ {0}).
Thus, (x,y) $\in$ ({0} x [2,3]) U ([2,3] x {0}).
Hence, (X x X) \ ( ({0} x {0}) U ([2,3] X [2,3]) ) $\subseteq$ ({0} x [2,3]) U ([2,3] x {0}).

This proves that (X x X) \ ( ({0} x {0}) U ([2,3] X [2,3]) ) = ({0} x [2,3]) U ([2,3] x {0}).


Now, by definition, both {0} x [2,3] and [2,3] x {0} are basis members for the product topology on X x X.
Hence, ({0} x [2,3]) U ([2,3] x {0}) is open in X x X (by definition of the basis for a topology).

Hence, ({0} x {0}) U ([2,3] X [2,3]) = (X x X) \ ( ({0} x [2,3]) U ([2,3] x {0}) ) is a closed subset of X x X with the product topology.