Give an example of infinite closed sub set of R that is not perfect.
My set theory is a bit rusty so here goes.
Ok, so we want a set that is closed but infinite, that is not perfect.
Closed says that all accumulation points are contained in the set. Not perfect means that there must be points in the set that are not accumulation points.
How about
$\displaystyle A\equiv\left\{ 1/n:\ n\in \mathbf{Z}^+ \right\} \cup \{0\}$
Define an open set to be any subset of $\displaystyle A$ containing 0 or is the empty set. You may need to prove that this collection of sets satisfies what it means to be open.
You can show that the set of accumulation points is $\displaystyle \{0\}$ which is a non-empty proper (strict) subset of $\displaystyle A$.
Part of the problem is that I don't know exactly from where you are starting. For example, different contexts have different definitions of open sets even though they may be equivalent in some cases. For example: Open set - Wikipedia, the free encyclopedia and Open set - Wikipedia, the free encyclopedia. So I'll give it a shot based on the 2 questions that I see you have posted.
A topological space, is a set, A, with a collection of subsets (which will be the collection of open subsets) that satisfy the following conditions:
- the empty set must be open
- The intersection of 2 (and thus finitely many) open sets must be open
- The arbitrary union (finite or infinite) of open sets must be open.
I found a problem with my original example so I am going to come up with a new (and hopefully correct) example.
Let $\displaystyle A=[0,1]$
Define the collection of open sets to be $\displaystyle \{\emptyset,\{0.5\}\}$.
The way I defined the collection of open sets, I explicitly declared the empty set to be open so #1 is satisfied.
#2. Prove by exhaustion:
$\displaystyle \emptyset \cup \{0.5\} = \emptyset$ which is open
#3. Prove by exhaustion again.
OK. we just proved that A is a valid topological space.
Now, we have to show that it is closed but not perfect. We know that A is infinite.
The closure of A is $\displaystyle [0,0.5) \cup (0.5,1]$. Proof: Let $\displaystyle x \in [0,0.5) \cup (0.5,1]$. Then $\displaystyle x$ is not contained in any open set. Thus, the following statement is trivially true: all open sets containing $\displaystyle x$ contain another point. Thus $\displaystyle x$ is a limit point. 0.5 is not a limit point since there is an open set, namely $\displaystyle \{0.5\}$, that does not contain another point.
Since $\displaystyle [0,0.5) \cup (0.5,1] \subsetneq [0,1]$ we know that $\displaystyle [0,1]$ is closed (defined with the above topology it contains all limit points) and that it is not perfect (since it contains a point that is not in the closure (not a limit point).