Regarding projection mapping Let $X$ be a topological space having Bolzano Weierstrass property and let $Y$ be a first countable space. I want to show that the projection mapping $p_Y:X\times Y\to Y$ is closed.
I tried as follows:
Let $C$ bea closed subset of $X\times Y$. We want to show that $p_Y(C)$ is closed. Let $x\in \overline{p_Y(C)}$. Since, $Y$ is first countable, therefore, there exists a sequence $(x_n)$ such that $x_n\to x$. I couldn't proceed further. Please give any hint.
 A: You need $X$ to be $T_1$. I can construct a counterexample with a $T_0$ space $X$; see the addition below.
Let $\{B_n:n\in\Bbb N\}$ be a countable base of open nbhds of $x$, and let $U=(X\times Y)\setminus C$. If $x\notin p_Y[C]$, then $U$ is an open nbhd of $\{x\}\times Y$, so for each $y\in Y$ there are an open nbhd $V_y$ of $y$ in $Y$ and an $n(y)\in\Bbb N$ such that $\langle x,y\rangle\in B_{n(y)}\times V_y\subseteq U$. For each $k\in\Bbb N$ let $$V_k=\bigcup\{V_y:y\in Y\text{ and }n(y)=k\}\;.$$


*

*Show that $B_k\times V_k\subseteq U$ for each $k\in\Bbb N$.  

*Verify that $\{x\}\times Y\subseteq\bigcup_{k\in\Bbb N}V_k$ and hence that $\{V_k:k\in\Bbb N\}$ is a countable open cover of $Y$.  

*Recall (or prove) that in $T_1$ spaces the property that every infinite set has a limit point is equivalent to countable compactness, and conclude that there is a finite $F\subseteq\Bbb N$ such that $\{V_k:k\in F\}$ covers $Y$.  

*Let $B=\bigcap_{k\in F}B_k$, and show that $\{x\}\times Y\subseteq B\times Y\subseteq U$.  

*Conclude that $x\notin\operatorname{cl}_Yp_Y[C]$.


Added: Let $X=\{x\in\Bbb R:x\ge 0\}$. For $x\in X$ let $U_x=\{y\in X:y<x\}$, and let $$\tau=\{X\}\cup\{U_x:x\in X\}\;;$$ $\tau$ is a $T_0$ topology on $X$. If $A\subseteq X$ is non-empty, let $x\in A$; then $x+1$ is a limit point of $A$, since every open nbhd of $x+1$ contains $x$.
Let $Y$ be $\{x\in\Bbb R:x\ge 0\}$ with the usual topology; certainly $Y$ is first countable. Let $$C=\{\langle x,y\rangle\in X\times y:xy\ge 1\}\;;$$ then $C$ is closed in $X\times Y$, but $p_Y[C]=Y\setminus\{0\}$ is not closed in $Y$.
