1. ## Cluster Point

let A be a subset of a space (X,T) and let $\displaystyle y\in A-A'$ A' is the derive set (the set that contains the clusters points ) show that {y} is open in the subspace $\displaystyle (A,T_A)$

I will try to show that A-{y} is closed by prove that $\displaystyle \overline{A-\left(y\right)}=A-\left(y\right)$
(y)={y} $\displaystyle \overline{A-\left(y\right)}$ is in $\displaystyle T_A$ not in (X,T) and it is enough to prove that $\displaystyle y\notin \overline{A-\left(y\right)}$

Suppose that $\displaystyle y\in \overline{A-(y)}$ $\displaystyle \Rightarrow \forall u_a \in T_A \text{ with } \text{ } y\in u_a\ : u\cap A \ne \phi$

but
$\displaystyle y\in A-A'\Rightarrow y\in A\cap (X-A')$ so

$\displaystyle y\in A\; and\; y\in (X-A')$

$\displaystyle y\in (X-A')\Rightarrow y\notin A' \Rightarrow \exists u\in T : u\cap A=\phi$

here where I stuck can anyone help me ??

Thanks

2. You will obtain the result from the existence of an open subset $\displaystyle U$ in $\displaystyle X$ containing $\displaystyle y$ such that $\displaystyle U\cap A=y$. But if such an open set does not exist, we would have a contradiction with $\displaystyle y\in A\setminus A'$.

3. Originally Posted by Enrique2
You will obtain the result from the existence of an open subset $\displaystyle U$ in $\displaystyle X$ containing $\displaystyle y$ such that $\displaystyle U\cap A=y$. But if such an open set does not exist, we would have a contradiction with $\displaystyle y\in A\setminus A'$.
Thanks very much

I think I can also solve it

Originally Posted by Amer
let A be a subset of a space (X,T) and let $\displaystyle y\in A-A'$ A' is the derive set (the set that contains the clusters points ) show that {y} is open in the subspace $\displaystyle (A,T_A)$

I will try to show that A-{y} is closed by prove that $\displaystyle \overline{A-\left(y\right)}=A-\left(y\right)$
(y)={y} $\displaystyle \overline{A-\left(y\right)}$ is in $\displaystyle T_A$ not in (X,T) and it is enough to prove that $\displaystyle y\notin \overline{A-\left(y\right)}$

Suppose that $\displaystyle y\in \overline{A-(y)}$ $\displaystyle \Rightarrow \forall u_a \in T_A \text{ with } \text{ } y\in u_a\ : u_a\cap A-(y) \ne \phi$

but
$\displaystyle y\in A-A'\Rightarrow y\in A\cap (X-A')$ so

$\displaystyle y\in A\; and\; y\in (X-A')$

$\displaystyle y\in (X-A')\Rightarrow y\notin A' \Rightarrow \exists u_0\in T : u_0\cap A-(y)=\phi$

let $\displaystyle v=u_0 \cap A$ and $\displaystyle y\in v$ so $\displaystyle v\in T_A$ and $\displaystyle v\cap A-(y)=\phi$ contradiction

I think it is now correct right ??

Thanks

4. Yes, you are right. In fact, proving that the complementary of y in A is closed is the same as showing that y is open. It seems that you are used to prove closedness instead of opennes.