Topology problems on closure of a set and limit points

Question

Need help here guys, In problem 2.26, I dont have an example which satisfies both the criteria. Each and every time, I am getting either of them but not both of them.

In example 2.30, I am not sure but I think open sets are going to be the set of natural numbers (symbolized as N) and closed sets will be {0} and empty set. But I dont know the limit points and other parameters.

In 2.32, I am completely clueless as to what its asking , because if x is a limit point of Y, then it implies that it can be a sequential limit point of Y. But here, its asking me the opposite.

I did this for 2.30, but my instructor told me to explain the points (ii) and (iii) while verifying that Ts is a topology. He told me that you will have to show the other ones too since its just not those elements but also its a finite sub-collection of elements of T for union and intersection. I don't know how many to show, since its an entire set of Natural nos.

Any help will be greatly appreciated. Thanks!!

Answer 1

Let $b$ be a bijection from $\mathbb N$ onto $\mathbb Q$ and consider the topology $\mathcal T$ on $\mathbb N$ defined by$$\mathcal T=\{S\subset\mathbb N\mid b(S)\text{ is open with respect to the usual topology on }\mathbb Q\}.$$Now, take $A=b^{-1}\bigl((-1,1)\bigr)$. Then $\overline A=b^{-1}\bigl([-1,1]\bigr)$.

The other problem has already been answered before.

Answer 2

2.26 Take the topology $\mathcal{T} = \{\Bbb N \setminus F: F \subseteq \Bbb N \text{ finite }, 0 \notin F\} \cup \{A \subseteq \Bbb N: 0 \notin A\}$. Then $A=\{2n+1: n \in \Bbb N\}$ is not closed, and it closure is $A \cup \{0\} \neq \Bbb N$.

2.30 The problem described you what the open sets are. The closed sets are their complements: i.e. all sets of the form $\{1,2,\ldots, N-1\}$ plus $\emptyset$ and $\Bbb N$ of course. If a set $A$ is infinite, all points of $\Bbb N$ are limit points of it. Think for yourself what limit points of finite sets can be, using the definition. As the space is first countable, all limit points will be sequential limit points as well.

2.32 Take $\Bbb R$ in the cocountable topology and $A = [0,1]$. Then $\pi$ is a limit point of $A$, but not a sequential limit point.

Answer 3

For problem 2.26, just go minimalist: You need a closed set that is neither empty nor the full space, thus you need an open set with that property (because closed sets are the complements of open sets). Therefore just define one such open set, say $\{0\}$. That is, the topology is $\mathcal T=\{\emptyset,\{0\},\mathbb N\}$. Then you just need to select a non-empty proper subset of that one non-trivial closed set, say $A=\{1\}$. Then $\overline A = \mathbb N\setminus \{0\}$ which clearly is neither $A$ nor $\mathbb N$.