Compact sets of Moore plane

Question

I'm trying to describe the compact sets of the Moore/Niemytzki plane.

One characteristic might be for the sets disjoint from the x-axis, if the set is closed and bounded, then it's compact. I guess if a set touched the x-axis in an infinite number of points, it would not be compact. So the compact sets consisting of a closed, bounded set disjoint from the x-axis along with finitely many points on the x-axis is compact?

Answer 1

If $C$ is compact in the Niemytzki plane $X$, then $C \cap D$ (where $D$ is the $x$-axis, which is closed and discrete in $X$) is a compact (closed subset of $C$) and discrete (subset of $D$) of $X$, so must be finite.

But not all compact subsets are of the form $C$ compact in $X\setminus D$ (the upper plane) plus a finite subset on $D$: consider a convergent sequence from the upper plane to a point on $D$, including its limit. The part outside $D$ is discrete, so non-compact.

I propose: all subsets that are compact in the usual topology on $X$ as a Euclidean subspace that have a finite intersection with $D$.

Answer 2

I will denote the Moore plane by $M$ and the $x$-axis (that is, $\Bbb R\times\{0\}$) by $X$. If $p\in\Bbb R^2$ and $r>0$, then $D_r(p)$ is the open disk centered at $p$ with radius $r$.

Let $K$ be a compact subset of $M$. Then $K$ is also compact with respect to the usual topology on $\Bbb R\times[0,\infty)$, since this topology is finer than the Moore plane topology. On the other hand, $K\cap X$ is a closed subset of $K$ (since $M$ is separated and $X$ is a closed subset of $M$), and therefore it is compact. But $X$ is a discrete subspace of $M$, and therefore its only compact subsets are the finite ones. It is then proved that $K$ is compact with respect to the usual topology and that $K\cap X$ is finite. Let $F_K=K\cap X$ and let $R_K$ be the set of the abscissas of the elements of $F_K$ (that is, if $x\in\Bbb R$, then $x\in R_K$ if and only if $(x,0)\in F_K$). Take $x\in R_k$ e let $x_1$ and $x_2$ be real numbers such that $x_1<x<x_2$ and that $[x_1,x_2]\cap R_K=\{x\}$; such numbers exist since $R_K$ is finite. Besides, $K\cap([x_1,x_2]\times[0,\infty))$ is a closed subset of $K$ and therefore it is compact. Take $r>0$ and consider the set \begin{equation}\label{eq:r8} D_r\bigl((x,r)\bigr)^\complement\cap K\cap([x_1,x_2]\times[0,\infty)).\tag1 \end{equation} Note that this set is compact and that its only element belonging to $X$ is $(x,0)$. Let us see that there is some $\varepsilon>0$ such that all points of \eqref{eq:r8} whose ordinate is greater than $0$ have, in fact, ordinate greater than or equal to $\varepsilon$. Otherwise, let us consider, for each $n\in\Bbb N$, the open subset $\Bbb R\times\left(\frac1n,\infty\right)$ of $M$. Then none of these open sets would contain the set of those elements of \eqref{eq:r8} whose ordinate is greater than $0$, and it would follow from this that \begin{equation}\label{eq:r9} \left\{\{(x,0)\}\cup D_r\bigl((x,r)\bigr)\right\}\cup\left\{\Bbb R\times\left(\frac1n,\infty\right)\,\middle|\,n\in\Bbb N\right\}\tag2 \end{equation} would be a set of open subsets of $M$ whose union would contain \eqref{eq:r8} and such that the union of any finite subset of \eqref{eq:r9} would not contain that set. This is impossible, since \eqref{eq:r8} is compact.

It is then proved that, if $K\subset M$, then, in order for $K$ to be compact it is necessary that

• $K$ is compact with respect to the usual topology;
• $F_K(=K\cap X)$ is finite;
• if $x\in\Bbb R$ is such that $(x,0)\in K$ and if $x_1,x_2,r\in\Bbb R$ are such that $x_1<x<x_2$, that $[x_1,x_2]\cap R_K=\{x\}$ and that $r>0$, then there is some $\varepsilon>0$ such that the ordinates of all points of \eqref{eq:r8} whose ordinate is greater than $0$ are greater than or equal to $\varepsilon$.

Now we will prove that these conditions are also sufficient to ensure that $K$ is a compact subspace of $M$. Let us then assume that these conditions hold.

We are then assuming that the set $F_K$ is finite. If it is empty, then $K\subset\Bbb R\times(0,\infty)$ and then to assert that it is compact as a subspace of $M$ is the same thing as asserting that it is compact as a subspace of $\Bbb R\times(0,\infty)$, and it is being assumed that this is true. Let us assume that $R_K$ is of the form $\{x_1,x_2,\ldots,x_n\}$, for some $n\in\Bbb N$, with $x_1,x_2,\ldots,x_n\in\Bbb R$ and $x_1<x_2<\cdots<x_n$. Take $y_0,y_n\in\Bbb R$ such that, for each $p\in K$, the abscissa of $p$ belongs to $[y_0,y_n]$ (these numbers exist, since $K$ is compact with respect to the usual topology, and therefore it is bounded) and let $y_1,y_2,\ldots,y_{n-1}\in\Bbb R$ be such that, for each $\in\{1,2,\ldots,n\}$, $y_{k-1}<x_k<y_k$. Then$$K=\bigcup_{k=1}^n\bigl(K\cap([y_{k-1},y_k]\times[0,\infty))\bigr),$$and therefore, in order to prove that $K$ is compact, it is enough to show that, for each $k\in\{1,2,\ldots,n\}$, $K\cap([y_{k-1},y_k]\times[0,\infty))$ is compact.

Take $k\in\{1,2,\ldots,n\}$ and let $\{A_\lambda\mid\lambda\in\Lambda\}$ be a set of open subsets of $M$ whose union contains $K\cap([y_{k-1},y_k]\times[0,\infty))$; we will show that there is a finite subset whose union still contains that set. Take $\lambda_0\in\Lambda$ such that $(x_k,0)\in A_{\lambda_0}$. Since $A_{\lambda_0}$ is an open set to which $(x_k,0)$ belongs, there is some $r>0$ such that $\{(x_k,0)\}\cup D_r\bigl((x_k,r)\bigr)\subset A_{\lambda_0}$. We are assuming that there is some $\varepsilon>0$ such that the ordinates of the points of$$D_r\bigl((x_k,r)\bigr)^\complement\cap K\cap([y_{k-1},y_k]\times[0,\infty))$$whose ordinate is greater than $0$ are actually greater than or equal to $\varepsilon$, that is, that they belong to $K\cap ([y_{k-1},y_k]\times[\varepsilon,\infty))$; let $K_\varepsilon$ be this set. Then $K_\varepsilon$ is compact and it is a subset of $K\cap([y_{k-1},y_k]\times[0,\infty))$; therefore, there are elements $\lambda_1,\ldots,\lambda_n$ of $\Lambda$ such that $K_\varepsilon\subset\bigcup_{j=1}^nA_{\lambda_j}$. On the other hand, the elements of $K\cap([y_{k-1},y_k]\times[0,\infty))$ whose ordinates are smaller than $\varepsilon$ must belong to $\{(x_k,0)\}\cup D_r\bigl( (x_k,r)\bigr)$, and therefore they belong to $A_{\lambda_0}$. It is then proved that$$K\cap([y_{k-1},y_k]\times[0,\infty))\subset\bigcup_{j=0}^nA_{\lambda_j}.$$

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As an example, let $K$ be the closed disk centered at $(0,1)$ with radius $1$. Is it a compact subset of the Moore plane? It is compact with respect to the usual topology and its intersection with the $x$-axis is finite; it consists of $\{(0,0)\}$. What about the third condition? It doesn't hold. Take $x=0$, $x_1=-2$, $x_2=2$, and $r=1$. Then$$D_r\bigl((0,r)\bigr)^\complement\cap K\cap([x_1,x_2]\times[0,\infty))=D_1\bigl((0,1)\bigr)^\complement\cap K\cap([-2,2]\times[0,\infty)),$$which is just the circle centered at $(0,1)$ with radius $1$. Clearly, it has points whose ordinates are greater than zero and arbitrarily close to $0$. Therefore, the third condition does not hold, and so $K$ is not a compact subset of the Moore plane. Of course, more generally no closed disk centered at a point $(x,r)$ with radius $r$ (with $x\in\Bbb R$ and $r>0$) is a compact subset of the Moore plane, and it follows from this that the Moore plane is not locally compact.

On the other hand, by this criterion, any triangle with one vertex on the $x$-axis and the other two vertices on $\Bbb R\times(0,\infty)$ is a compact subset of the Moore plane.