# Thread: Prove that the application is closed

1. ## Prove that the application is closed

Hi, everyone,
I resolved the following exercise. I need help reviewing it and giving the correct answer (if mine isn't correct of course).

Let's have:
$I_1=\{(x,y) \in R^2|y=0, 0\leq x\leq 1\}$
$I_2=\{(x,y)\in R^2|x=1, 0\leq y\leq 1\}$
$I_3=\{(x,y)\in R^2|y=x; x,y \in [0 \;1]\}$
$Q=I_1 \cup I_2 \cup I_3$

where the following equivalence relation is defined
$\A (x,y), (x',y') \in Q$, $(x,y)\sim (x',y')$ if $(x,y)=(x',y')$ or $(x, y), (x', y')\in I_1$,
$(x,y),(x',y')\in I_2$
Prove that $\pi : Q\rightarrow Q/\sim$ is closed.

Solution:

$Q$ is connected and compact so even $Q/\sim$ must be connected and compacted. At this point we have to prove that $Q/\sim$ is $T_2$ (Hausdorff). That is to prove that the set
$F= \{ (x;y),(u;v): \pi(x;y)=\pi(u;v) \}=\{ (x;y),(u;v): (x;y)\sim(u;v) \}$ is closed on $Q\times Q$.
$C_1=\{ (x;y),(x;y)\in Q\times Q| (x;y)\in Q}$, $C_2=\{ (x;0),(x;0)\in Q\times Q| x\in[0\;1]\}$
$C_3=\{ (1;y),(1;y)\in Q\times Q| y\in[0\;1]\}$.
$C_1$ is closed because $Q$ is $T_2$. $C_2$, $C_3$ are closed being subspaces of space $T_2$ and immages of the compact $[0\;1]$ on $R$ through continuous applications. Because $F=C_1 \cup C_2 \cup C_3$ then even $F$ will be closed, this means that $Q/\sim$ is $T_2$. I conclude that $\pi$ is closed.

2. ## Re: Prove that the application is closed

There are things I don't understant in your explanation. F, as defined, seems to be en element of Q/~ X Q/~ while C1,C2 and C3 are in Q X Q

3. ## Re: Prove that the application is closed

Sorry, but F is defined in $Q \times Q$, then I use the theorem that say: If F is closed in $Q\times Q$ then $Q/\sim$ is $T_2$ and this means that $\pi$ is closed.

4. ## Re: Prove that the application is closed

I think that $C_2\subset C_1$ and $C_3\subset C_1$ and threre are elements of F that are not in C1