Limit point(Cluster point) of the Singleton

Amer

Active member
Hey,
What is the limit point of the singleton ?
for example {x} in any topology
{x}' = phi
suppose not there exist x in {x}'
let U be an open set containing x, U should contain another point of {x} other than x but that impossible
did i do something wrong ?

Klaas van Aarsen

MHB Seeker
Staff member
Can you supply some additional information?

For starters what you mean by {x}' = phi?
What's the quote?
And what do you mean by phi?

Fantini

MHB Math Helper
I think that if your topology satisfies the $T_1$ or $T_2$ axioms we can conclude that it is itself, because it is going to be a closed set and therefore it contains all its limit points, but I am not sure.

My 2 cents. Amer

Active member
Can you supply some additional information?

For starters what you mean by {x}' = phi?
What's the quote?
And what do you mean by phi?
Ok
The question is we call a Topological space X a TD space, if for all $$x \in X$$
the derive set ( the set of all cluster points) of $$\{ x \}$$ is closed.
Prove that if a space is $$T_1$$ then it is a TD space and if a space is TD then it is $$T_0$$

so I tried to understand the TD space, I found that each space is TD space
$$(X,T)$$ be a topological space
let $$x \in X$$
I want to prove that the derive set of the Singleton is closed, I get that the derive set is $$\phi$$ which means the set $$\{x\}$$ dose not have any cluster point !! is that true ?

Amer

Active member
I found my mistake, in any topology $$(X,T)$$
the derive set of $$\{x\}$$ dose not contains x but it maybe contain another element for example
$$X = \{a,b\}$$ and $$T = \{X,\phi , \{a\} \}$$
the derive set of $$\{a\}$$ contains b
since each open set containing "b" contains "a" ( in fact we have just one open set containing b, X )

I am sorry because I am not clear in asking the question, but when I am writing I am thinking that all my thoughts are written or the reader has the same thoughts

Let $$(X,T)$$ be a TD space, and x,y in X
the set $$\{x \}'$$ is closed and dose not contain x
if $$y \in \{x\}'$$ , then $$U = X - \{x\}'$$ an open set which contains "x" and dose not contain y
if $$y$$ is not in $$\{x\}'$$ then there exist an open set $$U$$ containing "y" which dose not contains x ( the intersection of $$U- \{y\} \cap \{x\} = \phi$$ )
so the space is $$T_0$$ space.