# $(A^ctimes B)cup(Atimes B^c)cup(A^c times B^c)=(A^ctimes Y)cup(Xtimes B^c)$

Mathematics Asked by Yahya B. Jaber on January 5, 2022

Let $$(X,T)$$ and $$(Y,T’)$$ be two topological spaces and $$Asubset X ,Bsubset Y$$.Show that $$(A^ctimes B)cup(Atimes B^c)cup(A^c times B^c)=(A^ctimes Y)cup(Xtimes B^c)$$
solution $$1………..(A^ctimes B)cup(Atimes B^c)cup(A^c times B^c)=(A^ctimes B)cupBiggl((Acup A^c)times(B^ccup B^c)Biggl)$$
$$2……..=(A^ctimes B)cup(Xtimes (B^ccup B^c))$$
$$3…….=(A^ccup X times Bcup (B^ccup B^c))$$
$$4……=(A^ccup X times (Bcup B^c)cup B^c)$$
$$5……=(A^ccup X times Ycup B^c)$$
$$6….=(A^ctimes Y)cup(Xtimes B^c)$$
Is this proof correct?

Already in the first step you’ve written something that suggests a misconception on your part. It’s true that $$(Atimes B^c)cup(A^ctimes B^c)=(Acup A^c)times(B^ccup B^c)$$, but only because $$(Atimes B^c)cup(A^ctimes B^c)=(Acup A^c)times B^c$$; it is not in general true that

$$(Wtimes X)cup(Ytimes Z)=(Wcup Y)times(Xcup Z);.tag{1}$$

For instance, if $$W=X={0}$$ and $$Y=Z={1}$$, then

begin{align*} (Wtimes X)cup(Ytimes Z)&=({0}times{0})cup({1}times{1})\ &={langle 0,0rangle}cup{langle 1,1rangle}\ &={langle 0,0rangle,langle 1,1rangle};, end{align*}

but

begin{align*} (Wcup Y)times(Xcup Z)&=({0}cup{1})times({0}cup{1})\ &={0,1}times{0,1}\ &={langle 0,0rangle,langle 0,1rangle,langle 1,0rangle,langle 1,1rangle};. end{align*}

The remaining steps are ambiguous, since you didn’t use enough parentheses, but it appears that you really did think that $$(1)$$ is true and used it at your step $$3$$. There it definitely fails.

What is true is that $$(Xtimes Z)cup(Ytimes Z)=(Xcup Y)times Z;;tag{2}$$ you should prove this, if you’ve not done so already. Thus, a good first couple of steps would be

begin{align*} (A^ctimes B)cup(Atimes B^c)cup(A^ctimes B^c)&=(A^ctimes B)cupbig((Acup A^c)times B^cbig)\ &=(A^ctimes B)cup(Xtimes B^c);. end{align*}

Clearly $$(A^ctimes B)cup(Xtimes B^c)subseteq(A^ctimes Y)cup(Xtimes B^c)$$, so you could finish the proof by showing that $$(A^ctimes Y)cup(Xtimes B^c)subseteq(A^ctimes B)cup(Xtimes B^c)$$.

There are several ways to do this; for instance, you could use ‘element-chasing’, assuming that $$langle x,yranglein(A^ctimes Y)cup(Xtimes B^c)$$ and showing that $$langle x,yranglein(A^ctimes B)cup(Xtimes B^c)$$. In keeping with the more algebraic style that you were using, however, you could notice that

$$A^ctimes Y=A^ctimes(Bcup B^c)=(A^ctimes B)cup(A^ctimes B^c)$$

by $$(2)$$, so that

begin{align*} (A^ctimes Y)cup(Xtimes B^c)&=(A^ctimes B)cup(A^ctimes B^c)cup(Xtimes B^c)\ &=(A^ctimes B)cup(Xtimes B^c);, end{align*}

since $$A^ctimes B^csubseteq Xtimes B^c$$.

And as you can see, we’ve not just shown that $$(A^ctimes Y)cup(Xtimes B^c)subseteq(A^ctimes B)cup(Xtimes B^c);:$$ we’ve shown that $$(A^ctimes Y)cup(Xtimes B^c)=(A^ctimes B)cup(Xtimes B^c);.$$

If you wanted to, you could reverse the order of this last calculation and append it to the first part to get a direct proof of equality:

begin{align*} (A^ctimes B)cup(Atimes B^c)cup(A^ctimes B^c)&=(A^ctimes B)cupbig((Acup A^c)times B^cbig)\ &=(A^ctimes B)cup(Xtimes B^c)\ &=(A^ctimes B)cupbig((A^ctimes B^c)cup(Xtimes B^c)big)\ &=big((A^ctimes B)cup(A^ctimes B^c)big)cup(Xtimes B^c)\ &=(A^ctimes Y)cup(Xtimes B^c);. end{align*}

Answered by Brian M. Scott on January 5, 2022

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