$ KC $ spaces imply $ US $ spaces , but vise versa is false.

Question

In the $ US $ space , each convergent sequence has unique limit.

In the $ KC $ space , every compact subset is closed.

It easy to show that $ KC $ spaces imply $ US $ spaces. The example below show that $ US $ spaces do not imply $ KC $ spaces.

Let $ X = [ 0, 1 ] $ with the the usual topology and attach a new point z whose neighborhood are open dense subsets of $ [ 0,1 ]$. Observe $ [ 0,1 ]$ is a compact non-closed of $X$ and then $X$ is not a $KC$ space. However no sequence in $ [ 0,1 ]$ converge to $z$ and in particular all convergent in $X$ have unique limits.

can you give me more details?

1. Why $X$ is not closed?

2. Why no sequence in $ [ 0,1 ]$ converges to $z$ and in particular all convergent sequences in $X$ have unique limits ?

Answer 1

We define $X^*$ as the union $I\cup\{z\}$ such that a neighborhood base $\mathcal B_z$ of $z$ is given by those sets containing $z$ whose intersection with $I$ is an open dense subset of $I$, and a neighborhood base $\mathcal B_x$ of $x\in I$ is the usual neighborhood filter.
To see that this defines a topological space on $X^*$ via neighborhood filters, you have to show two things:

1. Each neighborhood base is indeed a filterbase, i.e. for all $A,B\in\mathcal B_x$ we have $A\cap B\supseteq C$ for some $C\in\mathcal B_x$.
2. For each $A\in\mathcal B_x$ there is a $B\in\mathcal B_x$ s.t. for each $y\in B$ there is a $C\in\mathcal B_y$ with $C\subseteq A$.

Both properties are easy to verify in this situation.

Let $\tau^*$ be the topology on $X^*$ and $\tau$ the topology on $I$. Note that $\tau^*$ induces the original topology on $I$. Therefore $I$ is still compact in $X^*$, but it is not closed since $\mathrm{cl}(I)=X^*$. So $X^*$ is not a $KC$ space.
Now let $(x_n)_n$ be a sequence in $X^*$ converging to a $z$ and to a point $y\in I$. Since $I$ is open, the sequence is eventually in $I$, so we can assume that it is contained entirely in $I$. Since $I$ is $T_2$, there can be no cluster point $x\ne y$ in $I$. Then $(x_n)_n\cup\{y\}$ is closed (in $I$) and nowhere dense, hence its complement is open and dense and thus constitutes a neighborhood of $z$. But this means that $(x_n)_n$ cannot converge to $z$ and, as already mentioned, its limit in $I$ is unique. So $X^*$ is a $US$-space.

Although, this does not answer your question on why no sequence in $I$ converges to $z$, it at least shows that there are $US$-spaces that are not $KC$-spaces.

Answer 2

1. $X=[0,1]\cup\{z\}$ is a closed subset of itself. What is not closed in $X$ is $[0,1]$: $z$ is obviously a limit point of $[0,1]$, since every open neighborhood of $z$ contains points of $[0.1]$, and $z\notin[0,1]$. In fact, $[0,1]$ is dense in $X$.

2. Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $[0,1]$. The relative topology on $[0,1]$ as a subspace of $X$ is the usual topology, so $\sigma$ has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ that converges to some $x\in[0,1]$. Let $F=\{x_{n_k}:k\in\Bbb N\}\cup\{x\}$; then $F$ is a closed, nowhere dense subset of $[0,1]$. Let $U=[0,1]\setminus F$; $U$ is a dense open subset of $[0,1]$, so $U\cup\{z\}$ is an open neighborhood of $z$ disjoint from $F$. It follows that $U\cup\{z\}$ does not contain any tail of the sequence $\sigma$, which therefore cannot converge to $z$. This shows that no sequence in $[0,1]$ converges to $z$. $[0,1]$ is $T_2$, so every convergent sequence in $[0,1]$ has a unique limit in $[0,1]$, and we've just shown that $z$ is not a limit of such a sequence, so every convergent sequence in $[0,1]$ has a unique limit. Finally, suppose that $\sigma=\langle x_n:n\in\Bbb N\rangle$ is any convergent sequence in $X$. We've just seen that if $\sigma$ has an infinite subsequence lying entirely in $[0,1]$, then $\sigma$ does not converge to $z$. Thus, if $\sigma$ has two limits, both must be in $[0,1]$; that's impossible, since $[0,1]$ is a $T_2$ subspace of $X$, so $\sigma$ must have a unique limit. If $\sigma$ does not have an infinite subsequence lying entirely in $[0,1]$, then there is an $m\in\Bbb N$ such that $x_n=z$ for all $n\ge m$, and clearly $z$ is the unique limit of $\sigma$. Thus, all convergent sequences in $X$ have unique limits, and $X$ is $US$.