# Thread: Qs in topology need to be solve !

1. ## Qs in topology need to be solve !

Q1:-
If ( X , T) & ( Y , J ) are both Hausdrff spaces , so is ( X x Y , P X x Y ) .

remark:- P X x Y = the topology of the basis of ( X x Y )

Q2:-
A subspace of a Hausdrff space is also a Hausdrff space .

Whith many many thnx

2. If $(x_1,y_1),(x_2,y_2)\in X\times Y$ pick open sets $U_1,U_2\subset X$ and $V_1,V_2\subset Y$ (all open in their respective topologies) such that $x_i\in U_i$ and $y_i\in V_i$ and $U_1\cap U_2=\emptyset = V_1\cap V_2$. What happens if $(U_1\times V_1) \cap (U_2\times V_2)\neq \emptyset$?

Let $x,y\in Y\subset X$ pick $U,V \subset X$ such that $x\in U$, $y\in V$ and $U\cap V= \emptyset$ then what about $U\cap Y$ and $V\cap Y$

3. Originally Posted by Jose27
If $(x_1,y_1),(x_2,y_2)\in X\times Y$ pick open sets $U_1,U_2\subset X$ and $V_1,V_2\subset Y$ (all open in their respective topologies) such that $x_i\in U_i$ and $y_i\in V_i$ and $U_1\cap U_2=\emptyset = V_1\cap V_2$. What happens if $(U_1\times V_1) \cap (U_2\times V_2)\neq \emptyset$?

Let $x,y\in Y\subset X$ pick $U,V \subset X$ such that $x\in U$, $y\in V$ and $U\cap V= \emptyset$ then what about $U\cap Y$ and $V\cap Y$
first of all :- many many thnks for your help
now
the first section is 100% true

but then ....???

Let $x,y\in Y\subset X$ pick $U,V \subset X$ such that $x\in U$, $y\in V$ and $U\cap V= \emptyset$ then what about $U\cap Y$ and $V\cap Y$[/QUOTE]

how came Y be a subset of X ??!!!!!

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now i think that , we begin with
let $(U_1\times V_1) \cap (U_2\times V_2)\ = h$

$h \in (U_1\times V_1) \$ & $h \in (U_2\times V_2)\$

then same how we will get a contradiction ( how i dont know ) ???

4. ok ive got the prove
and I`ll put it her soon