Topolgy

**Amer** · August 7th 2009, 11:14 PM

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

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

Proof:

$(\Rightarrow)$

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

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

thanks very much

**flyingsquirrel** · August 7th 2009, 11:48 PM

Hello,

Originally Posted by Amer

Proof:

$(\Rightarrow)$

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

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

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

**Amer** · August 8th 2009, 01:37 AM

Originally Posted by flyingsquirrel

Hello,

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

I think I can define the projection function from $P_i:\left(\prod_{k=0}^{n} X_k,T_p\right) \cong \left(X_i,T_i\right)$
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

**flyingsquirrel** · August 8th 2009, 02:07 AM

Originally Posted by Amer

I think I can define the projection function from $P_i:\left(\prod_{k=0}^{n} X_k,T_p\right) \cong \left(X_i,T_i\right)$
this is one-one and onto and the projection is continuous and the inverse for the projection is continuous thus it is homeomorphism right

$P_i$ is onto and continuous but it not necessarily one-to-one. For example let

$\begin{aligned}P_x : \mathbb{R}\times \mathbb{R}& \to \mathbb{R}\\ (x,y)& \mapsto x \end{aligned}.$

We have $P_x((1,0))=P_x((1,2))=1$ so $P_x$ is not injective.

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

**Amer** · August 8th 2009, 02:15 AM

Originally Posted by flyingsquirrel

$P_i$ is onto and continuous but it not necessarily one-to-one. For example let

$\begin{aligned}P_x : \mathbb{R}\times \mathbb{R}& \to \mathbb{R}\\ (x,y)& \mapsto x \end{aligned}.$

We have $P_x((1,0))=P_x((1,2))=1$ so $P_x$ is not injective.

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

Thanks I get it now

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