Each factor space is homeomorphic to the subspace of the product Suppose that $\prod X_{\alpha}$ is a nonempty product space:
a)I need to prove  that if V is a nonempty open subset of  $\prod X_{\alpha}$ then  $\pi_{\alpha} (V)=X_{\alpha}$ for all but finitely many $\alpha$
now V is equal to  $\bigcup{\prod U_{\alpha}}$,  $U_{\alpha}$ is open in  $ X_{\alpha}$ and $U_{\alpha}$ =$ X_{\alpha}$ for all but finitely coordinates, so $\pi_{\alpha} (V)=\pi_{\alpha} \left(\bigcup{\prod U_{\alpha}}\right)=X_{\alpha}$ for all but finitely many $\alpha$ since the union of $ X_{\alpha}$ is equal to $ X_{\alpha}$ or some $\bigcup U_{\alpha}$ . IS this correct?!
and
b) If $c_{\alpha}\in X_{\alpha}$ for each ${\alpha}\in A$, then for each $\beta, X_{\beta}$ is homeomorphic to the subspace 
$$X_{\beta}^*=(x \in \prod X_{\alpha}:x_{\alpha}=c_{\alpha} \forall {\alpha}\neq \beta)$$
hence I need to show that: Each factor space is homeomorphic to the subspace of the product.
I have been trying to see how the elements of $X_{\beta}^*$ look like and I'm still fuzzy. I really need help.How is $X_{\beta}^*$ any different from $\prod X_{\alpha}$?
 A: (a) You could make this a bit clearer: you haven’t really explained why $\pi_\alpha\left(\prod_\alpha U_\alpha\right)=X_\alpha$ for all but finitely many $\alpha$. I would do it like this. Let $A$ be the set of indices. Since $U\ne\varnothing$, there is a basic open set $B=\prod_{\alpha\in A}U_\alpha\subseteq U$, where $U_\alpha$ is a non-empty open set in $X_\alpha$ for each $\alpha\in A$, and $F=\{\alpha\in A:U_\alpha\ne X_\alpha\}$ is finite. Clearly $\pi_\alpha[U]\supseteq\pi_\alpha[B]=U_\alpha$ for each $\alpha\in A$, so for each $\alpha\in A\setminus F$ we have $X_\alpha\supseteq\pi_\alpha[U]\supseteq\pi_\alpha[B]=U_\alpha=X_\alpha$. That is, $\pi_\alpha[U]=X_\alpha$ for every $\alpha\in A\setminus F$, so $\{\alpha\in A:\pi[U]\ne X_\alpha\}\subseteq F$ and is therefore finite.
(I think that you understood what was happening, but at this stage that’s only half the battle: the rest is figuring out how to express that understanding clearly.)
(b) Fix points $c_\alpha\in X_\alpha$ for each $\alpha\in A$, and for each $\beta\in A$ let
$$X_\beta^*=\left\{x\in\prod_{\alpha\in A}X_\alpha:x_\alpha=c_\alpha\text{ for all }\alpha\in A\setminus\{\beta\}\right\}\;.\tag{1}$$
I’ll explain what’s going on and give you a push in the right direction.
To see what $X_\beta^*$ looks like, let’s consider a very simple example. Let $A=\{1,2,3\}$, and for $k\in A$ let $X_k=\Bbb R$. Then $\prod_{k\in A}X_k=\prod_{k=1}^3X_k=\Bbb R^3$. Let $c_1=2$ and $c_2=1$; what is $X_3^*$?
$$X_3^*=\left\{\langle x,y,z\rangle\in\Bbb R^3:x=2\text{ and }y=1\right\}\;;$$
geometrically, $X_3^*$ is the vertical line through the point $\langle 2,1,0\rangle$ in the $xy$-plane. You’ve pinned down all but the third coordinate, and you’re letting that one range freely over $\Bbb R$. That line is clearly homeomorphic to $\Bbb R$ by the map $\langle 2,1,z\rangle\mapsto z$, which is just the projection $\pi_3$ restricted to $X_3^*$.
The same thing happens in general. The definition of $X_\beta^*$ in $(1)$ pins down every coordinate except the $\beta$-th coordinate to a fixed value and lets the $\beta$-th coordinate range freely over $X_\beta$. In very rough pictorial terms, $X_\beta^*$ is like a ‘line’ parallel to the $\beta$ axis. To show that $X_\beta^*$ is homeomorphic to $X_\beta$, try the same idea: show that the map $\pi_\beta\upharpoonright X_\beta^*:X_\beta^*\to X_\beta$, which just picks off the $\beta$-th coordinate of each point in $X_\beta^*$, is a homeomorphism.
