# Complements on the product topology

i'm working with the product topology of two topological spaces $X$ and $Y$ and i came across with the question to find closed sets in $X\times Y$, i know that the closed sets are the complement of the open sets but supose we have a basic open set in this product space, let $U \times V$ be it where $U$ are open in $X$ and $V$ open in $Y$, can i say that the complement of this set is $U^c\times V^c$?

There is a exercise on Mankres that he say if $X\times V$ are open so $X \times (Y-V)$ are closed, anyone can explain that too?

• Regarding your question "can i say that the complement of $U \times V$ is $U^c \times V^c$", ask yourself some related questions: Given $(x,y) \in X \times Y$, what does it mean that $(x,y) \in U \times V$? How do you negate that property? – Lee Mosher Jun 19 '17 at 21:40

Not quite. From de Morgan's law, $\neg(u\wedge v)=(\neg u)\vee(\neg v)$, we get $(U\times V)^c=(U^c\times Y)\cup (X\times Y^c)$. You can also write this as $(U^c\times V)\cup(U\times V^c)\cup(U^c\times V^c)$, which has more terms but may sometimes be easier to work with.

Hint. If $X$ and $Y$ are sets and $A \subseteq X$, $B \subseteq Y$ we have the formula $$(A \times B)^c = (A^c \times Y) \cup (X \times B^c)$$ Try to prove it.

$X \times V \subset X \times Y$ open implies $V$ open in $Y$ (projections are open maps) so $Y - V$ is closed in $Y$, so $X \times (Y-V)$ is closed in $X \times Y$ (product of closed sets is closed).

Or even easier $X \times (Y-V) = (X \times Y) - (X \times V)$ and so is a complement of a (basic) open set, hence closed, as $(x,y) \in X \times (Y-V)$ iff $x \in X, y \in Y - V$ iff $x \in X, y \in Y, y \notin V$ iff $(x,y) \in X \times Y, (x,y) \notin (X,V)$.

Since the question about $U^c\times V^c$ is answered I answer just the other ones.

The projections $$p_X:X\times Y\to X,~p_X(x,y)=x\text{ and }p_Y:X\times Y\to Y,~p_Y(x,y)=y$$ are continuous in the product topology. If $A\subset X$ closed, $U\subset X$ open, $B\subset Y$ closed and $V\subset Y$ open, then by continuoity of $p_X$ and $p_Y$ you get $$X\times V=p_Y^{-1}(V),~U\times Y=p_X^{-1}(U), U\times V=p_X^{-1}(U)\cap p_Y^{-1}(V)$$ are open. The same way are $$X\times B=p_Y^{-1}(B),~A\times Y=p_X^{-1}(A),~A\times B=p_X^{-1}(A)\cap p_Y^{-1}(B)$$ closed.

The product of open sets in $X$ and $Y$ yields an open set in $X\times Y$ and the product of closed sets in $X$ and $Y$ yields a closed set in $X\times Y$.

That are obviously not all open and closed sets. You can unite an arbitrary number of open sets to an open set or intersect finite many open set to an open set. The same way you can unite finite many closed sets to a closed set and intersect arbitrary many closed sets to a closed set.

Your last question is why $X\times(Y\setminus V)$ is closed while $X\times V$ is open. That holds since $$X\times (Y\setminus V)=(X\times Y)\setminus(X\times V)=(X\times V)^c.$$