- #1
sa1988
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Homework Statement
Prove the product of two Hausdorff spaces is Hausdorff
Homework Equations
The Attempt at a Solution
For ##X##, ##Y## Hausdorff spaces, need to find ##s, t \in X \times Y## where neighbourhoods ##S, T## of ##s## and ##t## are disjoint.
Firstly, by the definition of the Hausdorff property, ##\exists \ x_1, x_2 \in X## whose neighbourhoods ##U_1## and ##U_2## are such that ##U_1 \cap U_2 = \emptyset##,
and the same can be said for some ##y_1, y_2 \in Y##.
The product ##X \times Y## must therefore contain points ##(x_1, y_1), (x_2, y_2)## made of these elements, whose neighbourhoods are the sets:
##T = \{ (x_1 - \delta , x_1 + \delta), (y_1 - \delta , y_1 + \delta) \}## with ##0 < \delta < min(x_1,y_1)##
and
##S = \{ (x_1 - \epsilon , x_2 + \epsilon), (y_2 - \epsilon, y_2 + \epsilon) \}## with ##0 < \epsilon < min(x_2, y_2)##
Now by original construction, ##x_1 \not= x_2## and ##y_1 \not= y_2##, so it can be said that ##x_1 < x_2## and ##y_1 < y_2## without loss of generality.
and by the original condition for the Hausdorff spaces ##X## and ##Y##, the points ##x_1##, ##x_2##, ##y_1##, and ##y_2## all have open and separate neighbourhoods in their respective spaces, so there must exist ##\delta## and ##\epsilon## such that
##x_1 + \delta < x_2 - \epsilon##
and
##y_1 + \delta < y_2 - \epsilon##
which means there are two points ##t = (x_1, y_1)## and ##s = (x_2, y_2) ## in ##X \times Y## such that
##(x_1 + \delta , y_1 + \delta) < (x_2 - \epsilon , y_2 - \epsilon)##
meaning ##t < s \ \forall \ s, t## under the above construction,
whose neighbourhoods ##T## and ##S## cannot intersect.
Hence the product of two Hausdorff spaces is Hausdorff.Well... that's my effort. Not sure how solid it is though. Feels like I invoked a circular argument along the way...
Thanks.