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  1. Why if $\{A_\alpha\}_{\alpha \in \Omega}$ is a collection of path connected spaces. We must show that $\cup_{\alpha \in \Omega}$ is path connected.

    • I think $\cap_{\alpha \in \Omega}A_\alpha$ must be nonempty, but I can't prove $\cup_{\alpha \in \Omega}$ is path connected.
  2. Prove that $P_\beta : \prod_\alpha X_\alpha \to X_\beta$ is continuous, open and onto for all $\beta.$

    • I can prove $P_\beta$ is continuous and open. But I can't proof $P_\beta$ is onto [I think it's easy.] Please hint me to get $P_\beta$ is onto
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  1. Let $z \in \bigcap_{\alpha\in\Omega} A_\alpha$ and let $x,y \in \bigcup_{\alpha\in\Omega} A_\alpha$. Why is there a path $\gamma_1$ from $x$ to $z$? Why is there a path $\gamma_2$ from $y$ to $z$? Can you make a new path $\gamma$ from $x$ to $y$ that combines $\gamma_1$ and $\gamma_2$ in some way?
  2. What is the definition of $P_\beta$? If I have some $x_\beta \in X_\beta$ and I choose arbitrary points $x_\alpha \in X_\alpha$ for $\alpha \ne \beta$, what is $P_\beta((x_\alpha)_\alpha)$?
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Please check the proof 1. Let $z \in \cap_{\alpha \in \Omega}A_\alpha$ and let $x,y \in \cup_{\alpha \in \Omega}A_\alpha.$ There exist $\alpha,\beta \in \Omega$ such that $$x \in A_\alpha, y \in A_\beta.$$ Since $z \in \cap_{\alpha \in \Omega}A_\alpha$, we have $$z \in A_\alpha, z \in A_\beta.$$ Since $A_\alpha$ is path connected for all $\alpha,$ we have there are path $f_1 : [a,b] \to A_\alpha$ from $x$ to $z$ and path $f_2 : [c,d] \to A_\alpha$ from $y$ to $z$ such that $$f_1(a) = x, f_1(b) = z , f_2(c) = y, f_2(d) = z.$$ How I define new path from $x$ to $y$. ?? – Tee Mth Nov 20 '12 at 4:52
@TeeMth: Try $f:[a,b+d-c] \rightarrow \bigcup_{\alpha\in\Omega} A_\alpha$ given by $f(t)=f_1(t)$ for $t \in [a,b]$ and $f(t)=f_2(b+d-t)$ for $t \in [b,b+d-c]$. – wj32 Nov 20 '12 at 5:50
Thank you very much. Please hint me some problems In First Countable space: F is closed iff $\forall x_n \subset F$ and $x_n \to x$, then $x \in F$ Proof. Suppose that $\forall x_n \subset F$ and $x_n \to x$, then $x \in F$. We will show that $F$ is closed. That is show that $F = \bar{F}.$ Since $F \subset \bar{F},$ I sufficient to show that $\bar{F} \subset F.$ Let $x \in \bar{F}.$ Note that in First countable space, $x \in \bar{F}$ iff $\exists \{x_n\}$ in $F$ such that $x_n \to x.$ By assumption, we have $x \in F.$ Please you check the proof. $(\rightarrow)$I can't proof – Tee Mth Nov 20 '12 at 7:02
@TeeMth: If you find this answer helpful, please accept it. You should create a new question about first countable spaces. – wj32 Nov 20 '12 at 8:09

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