Example of $US$ space that is not $KC$ space.

Question

An example of a $US$ space that is not a $KC$ space is as follows.

Consider the space $[0,1]$ with the usual topology $\tau_u$.

Let $X=[0,1]\cup \{2\}$ and $\tau=\tau_u \cup \{U\cup \{2\} : U\in \tau_u$ and $U$ is a dense subset of $[0,1]\}$ the topology of $X$. I already showed that $\tau$ is a topology in X.

Now, I want to prove that $(X, \tau)$ is a $US$ space. My attempt is as follows

Let $\{x_n\}_{n \in \mathbb{N}}$ be a sequence in $X$ that converges to $x\in X$. Suppose that for an infinite subset $I$ of $\mathbb{N}$ it happens that $x_n=2$ for all $n\in I$. If $x\in [0,1]$ then, since $[0,1]$ is an element of $\tau_u$, all the elements of the sequence $\{x_n\}_{n \in \mathbb{N}}$ are in $[0,1]$, except for a finite quantity. This contradicts the choice of $I$, so $x=2$. If the sequence $\{x_n\}_{n \in \mathbb{N}}$ converges to $y\in X-\{x\}$, then $y\in [0,1]$ and repeating the previous argument, we have a contradiction to the fact that an infinity of members of $\{x_n\}_{n \in \mathbb{N}}$ are 2, we conclude that the sequence $\{x_n\}_{n \in \mathbb{N}}$ only converges to $x = 2$.

Now suppose that a finite number of elements of $\{x_n\}_{n \in \mathbb{N}}$ are 2. Let $J\subset N$ be such that $\{x_n\}_{n \in J}\subset [0,1]$. If $x=2$ we define

$U=[0,1]-\{x_n : n\in J \}$ and $V=U\cup \{2\}$

My problems are:

1.- Prove that $U\in \tau_u$ and dense subset of $[0,1]$

2.- What happens if $x_n\neq 2$?

Can you give me more details?

Answer 1

Your proof for the case that infinitely many $x_n = 2$ is correct, but can be shortened. You have shown that the assumption $x_n \to x \in [0,1]$ leads to a contradiction which means that necessarily $x = 2$. That is, the limit of $(x_n)$ is unique. No further steps are required here.

You have problems with the case that only finitely many $x_n = 2$. In this case we may w.l.o.g. asssume that all $x_n \in [0,1]$ because thee omission of finitely many elements of the sequence does not influence the convergence of $(x_n)$.

Note that the subspace topology on $[0,1] \subset X$ is nothing else than $\tau_u$.

1. $(x_n)$ does not converge to $2$.

Since all $x_n \in [0,1]$ and $[0,1]$ is compact, there exists a subsequence $(x_{n_k})$ converging to some $\xi \in [0,1]$. Since $(x_n)$ and $(x_{n_k})$ have the same limits, we may assume w.l.o.g. that $x_n \to \xi$. The set $A = \{x_n \mid n \in \mathbb N\} \cup \{\xi\}$ is a closed countable subset of $[0,1]$, and thus does not have interior points in $[0,1]$. Hence $U = [0,1] \setminus A$ is open and dense in $[0,1]$.

Now assume that $x_n \to 2$. Then we must have all but finitely many $x_n \in U \cup \{2\}$ which is a contradiction.

2. $(x_n)$ has a unique limit.

We know that all possible limits must be contained in $[0,1]$. But $[0,1]$ is Hausdorff, thus the limit is unique.