Topology - Compactness of $\mathbb{Z}\times\{0,1\}$ A question from my h.w.:


  Is the topological space $\mathbb{Z}\times\{0,1\}$
    (where $\mathbb{Z}$
    has the discrete topology and $\{0,1\}$
    the trivial one) compact? sequentially compact? or compact at accumulation points?


I'm not sure I understand the construction, but I don't see how this space has any notion of compactness - the open covering $\{z\}\times\{0,1\}_{z\in\mathbb{Z}}$
 doesn't have a finite sub covering, same goes for the sequence $\{z\times\{0\}\}_{z\in\mathbb{Z}}$
 that doesn't have a limit or an accumulation point. Did I misunderstand something?

Thanks for the help!
Well, you already got the answer but adding my point of view won't do any harm :)

Denote $X=\mathbb{Z}\times\{0,1\}$.


The space X is not compact: as you noticed $\{z\}\times \{0,1\}$ does not have a finite subcover.
The space X is not sequentially compact: consider the sequence $(a_n)_{n=1}^\infty$, $a_n=(n,1)$. 

No subsequence of $a_n$ converges $x_0\in X$, for if such $(a_{n_j})_{j=1}^\infty$ exists with $lim_{j\to\infty}a_{n_j}=x_0$,

any open neighbourhood of $x_0$ (denoted $U$) has elements of the tail $(a_{n_j})_{j=l}^\infty$ in it, for any $l\in \mathbb{N}$.

In other words: $\forall l\in\mathbb{N},\forall U_{x_0}: U_{x_0}\cap\{a_{n_j}|j&gt;l\}\neq\phi$.

But looking at $U_{x_0}=\{z\}\times\{0,1\}$ (where z is the first coordinate of $x_0$) and $l=z+1$ give the contradiction, hence X is not sequentially compact.
The space X is limit point compact: Let look at a more common definition of limit point compactness: X is said to be limit point compact if every infinite subset of X has a limit point in X, i.e a point x for which every neighbourhood of x contains at least one point of X different from x itself.

Let $S\subset X$ be an infinite set.
$i,j\in\{0,1\}$ and $i\neq j$.
Let $y=(k,i)\in S$ then any open neighbourhood of $x=(k,j)$ (even the smallest one $\{k\}\times \{0,1\}$) contains $y\in S$.
Well, you already got the answer but adding my point of view won't do any harm :)

Denote $X=\mathbb{Z}\times\{0,1\}$.


The space X is not compact: as you noticed $\{z\}\times \{0,1\}$ does not have a finite subcover.
The space X is not sequentially compact: consider the sequence $(a_n)_{n=1}^\infty$, $a_n=(n,1)$. 

No subsequence of $a_n$ converges $x_0\in X$, for if such $(a_{n_j})_{j=1}^\infty$ exists with $lim_{j\to\infty}a_{n_j}=x_0$,

any open neighbourhood of $x_0$ (denoted $U$) has elements of the tail $(a_{n_j})_{j=l}^\infty$ in it, for any $l\in \mathbb{N}$.

In other words: $\forall l\in\mathbb{N},\forall U_{x_0}: U_{x_0}\cap\{a_{n_j}|j&gt;l\}\neq\phi$.

But looking at $U_{x_0}=\{z\}\times\{0,1\}$ (where z is the first coordinate of $x_0$) and $l=z+1$ give the contradiction, hence X is not sequentially compact.
The space X is limit point compact: Let look at a more common definition of limit point compactness: X is said to be limit point compact if every infinite subset of X has a limit point in X, i.e a point x for which every neighbourhood of x contains at least one point of X different from x itself.

Let $S\subset X$ be an infinite set.
$i,j\in\{0,1\}$ and $i\neq j$.
Let $y=(k,i)\in S$ then any open neighbourhood of $x=(k,j)$ (even the smallest one $\{k\}\times \{0,1\}$) contains $y\in S$.