1. ## Topolgy

to prove that $\displaystyle T_0-space$ is a productive property

a product topology $\displaystyle \left(\prod_{k=1}^{n} X_k , T\right)$ is a $\displaystyle T_0$ space $\displaystyle \Longleftrightarrow$ for every k=1,...,n $\displaystyle (X_k , T_k)$ is a $\displaystyle T_0$ space

Proof:

$\displaystyle (\Rightarrow)$

let $\displaystyle \left(\prod_{k=1}^{n} X_k\right)$ is a T(0) space , let $\displaystyle 1\leq i \leq n$ then $\displaystyle \left(X_i , T_i \right)$ is homeomorphic to a subspace $\displaystyle Y_i$ of the product space $\displaystyle \prod_{k=1}^{n} X_k$

my question is how we know that there exist a subspace $\displaystyle Y_i$ which is homeomorphic to $\displaystyle (X_i,T_i)$ ?/

thanks very much

2. Hello,
Originally Posted by Amer
Proof:

$\displaystyle (\Rightarrow)$

let $\displaystyle \left(\prod_{k=1}^{n} X_k\right)$ is a T(0) space , let $\displaystyle 1\leq i \leq n$ then $\displaystyle \left(X_i , T_i \right)$ is homeomorphic to a subspace $\displaystyle Y_i$ of the product space $\displaystyle \prod_{k=1}^{n} X_k$

my question is how we know that there exist a subspace $\displaystyle Y_i$ which is homeomorphic to $\displaystyle (X_i,T_i)$ ?/
Let $\displaystyle (a_1,a_2,\ldots,a_{i-1},a_{i+1},\ldots,a_n)\in \prod_{\substack{k=1\\k\neq i}}^n X_k$. Then $\displaystyle Y_i=\{a_1\}\times \{a_2\}\times \cdots\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots \times \{a_n\}$ (with the topology induced by the topology on $\displaystyle \prod_{k=1}^{n} X_k$) is homeomorphic to $\displaystyle X_i$ because the map $\displaystyle \pi:Y_i \to X_i$ defined by $\displaystyle \pi\left((a_1,\ldots,a_{i-1},x_i,a_{i+1},\ldots, a_n)\right)= x_i$ is an homeomorphism. Can you show it ?

3. Originally Posted by flyingsquirrel
Hello,

Let $\displaystyle (a_1,a_2,\ldots,a_{i-1},a_{i+1},\ldots,a_n)\in \prod_{\substack{k=1\\k\neq i}}^n X_k$. Then $\displaystyle Y_i=\{a_1\}\times \{a_2\}\times \cdots\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots \times \{a_n\}$ (with the topology induced by the topology on $\displaystyle \prod_{k=1}^{n} X_k$) is homeomorphic to $\displaystyle X_i$ because the map $\displaystyle \pi:Y_i \to X_i$ defined by $\displaystyle \pi\left((a_1,\ldots,a_{i-1},x_i,a_{i+1},\ldots, a_n)\right)= x_i$ is an homeomorphism. Can you show it ?
I think I can define the projection function from $\displaystyle P_i:\left(\prod_{k=0}^{n} X_k,T_p\right) \cong \left(X_i,T_i\right)$
this is one-one and onto and the projection is continuous and the inverse for the projection is continuous thus it is homeomorphism right

Thanks

4. Originally Posted by Amer
I think I can define the projection function from $\displaystyle P_i:\left(\prod_{k=0}^{n} X_k,T_p\right) \cong \left(X_i,T_i\right)$
this is one-one and onto and the projection is continuous and the inverse for the projection is continuous thus it is homeomorphism right
$\displaystyle P_i$ is onto and continuous but it not necessarily one-to-one. For example let
\displaystyle \begin{aligned}P_x : \mathbb{R}\times \mathbb{R}& \to \mathbb{R}\\ (x,y)& \mapsto x \end{aligned}.
We have $\displaystyle P_x((1,0))=P_x((1,2))=1$ so $\displaystyle P_x$ is not injective.

However the restriction of $\displaystyle P_i$ to the set $\displaystyle Y_i=\{a_1\}\times\cdots\times\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots\times \{a_n\}$ that I defined in my previous post is an homeomorphism... it is the function I named $\displaystyle \pi$.

5. Originally Posted by flyingsquirrel
$\displaystyle P_i$ is onto and continuous but it not necessarily one-to-one. For example let
\displaystyle \begin{aligned}P_x : \mathbb{R}\times \mathbb{R}& \to \mathbb{R}\\ (x,y)& \mapsto x \end{aligned}.
We have $\displaystyle P_x((1,0))=P_x((1,2))=1$ so $\displaystyle P_x$ is not injective.

However the restriction of $\displaystyle P_i$ to the set $\displaystyle Y_i=\{a_1\}\times\cdots\times\{a_{i-1}\}\times X_i\times \{a_{i+1}\}\times \cdots \{a_n\}$ that I defined in my previous post is an homeomorphism... it is the function I named $\displaystyle \pi$.
Thanks I get it now