# How to prove such a space is Hausdorff and compact?

I am trying to prove the following statement.

Consider $C_{i} = \{(x,y):x^2+y^2=i^2\},i=1,2$ two circle in $\mathbb{R}^2$ plane, and define local bases as following:

• $x\in C_{2},\quad \mathcal{B}_x = \{\{x\}\}$, isolated
• $x\in C_{1}$, say $x=e^{it}$, let $$B(x,n) = \{e^{is}:|s-t|<2^{-n}\}\cup\{2e^{is}:0<|s-t|<2^{-n}\}$$ e.g. two archs on two circles, and $\mathcal{B}_x=\{B(x,n):n\in\mathbb{N}\}.$

Claim: the space is compact and Hausdorff.

Hausdorff is relatively easy to prove, the only difficulty is to prove the case a point $a=e^{ip}\in C_{1}$ and $b=2e^{ip}\in C_{2}$ , how can I construct such two disjoint open sets containing one of each?

To prove compactness, I think it should be prove by contradiction. Or can it be proved by the theorem:

if $f:X\rightarrow Y$ is a continuous and surjective map, where $X$ is compact, then $Y$ is compact too.

Though I have troubles to construct such a map.

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This looks like the Alexandrov double circle (from Engelking's book), but the description seems off. Are you sure you have the definition of $\mathcal{B}_x = B(x,n)$ right? Don't you want to remove the point $2e^{it}$ from it? Otherwise I don't think $e^{it}$ and $2e^{it}$ have disjoint neighborhoods. To prove compactness, use Alexander's subbase lemma and observe that $C_1$ carries its usual (compact) topology from $\mathbb{R}^2$. – t.b. Apr 29 '12 at 23:15
@t.b. Thanks a lot. I have overlooked the definition. Now it makes sense now. – newbie Apr 29 '12 at 23:26
Accidentally typed my comment in the answer box again :S Can you please change it to read "and that $C_1$ is compact". The whole space is obviously noncompact, what with all the isolated points in $C_2$. The posted excerpt makes this clear too. – rschwieb Apr 30 '12 at 12:40
@rschwieb: you're wrong, the space $C_1 \cup C_2$ is compact. The reason you mention is no reason for compactness to fail. – t.b. Apr 30 '12 at 12:43
@t.b. Hm, I must not be understanding the topology. It just seems nonsensical that $C_2$ can be discrete and the whole space can be compact. Since the excerpt made no claim that the whole thing is compact, I was tempted to believe it wasn't. Is the set $C_1$ together with the sets $\{z\}$ such that $z\in C_2$ not an open cover of the space? – rschwieb Apr 30 '12 at 12:44

After you fixed the definition of $B(x,n)$, the Hausdorff property should be easy to check. Distinguish four cases: $x,y \in C_1$, $x \in C_1$ and $y \in C_2$ do [or do not] lie on the same ray emanating from the origin and $x,y \in C_2$.

Observe that the circle $C_1 \subset X$ carries its usual topology inherited from $\mathbb{R}^2$.

To check compactness of $X = C_1 \cup C_2$ I would use Alexander's subbase lemma (see also here for a proof), which tells us that it suffices to consider covers $\{U_{i}\}_{i \in I}$ consisting of sets from the subbase $\bigcup_{x \in X} \mathcal{B}_{x}$ of the topology of $X$.

Since $C_1$ has its usual compact topology, we can find $x_{1},\ldots,x_{n} \in C_1$ and $U_{i_k} = B(x_{k},n_k)$ such that $C_1 \subset U_{i_1} \cup \cdots \cup U_{i_n}$. But then all of $C_1 \cup C_2$ is covered, except possibly the points $2x_{k}$. Since we started with a cover of $C_{1} \cup C_{2}$, we can find $V_{k}$ containing $2 x_{k}$ from the cover $\{U_{i}\}_{i \in I}$ we started with. Then $\{U_{i_1},\dots,U_{i_n},V_{1},\dots,V_{n}\}$ covers $C_1 \cup C_2$ and is a subcover of $\{U_{i}\}_{i \in I}$. Thus we conclude from Alexander's lemma that $X$ is compact.

As I mentioned in a comment, this space $X$ is often called Alexandrov's double circle and appears e.g. as example 3.1.26 in Engelking's General topology (here are two screen shots containing the relevant excerpt: part 1 and part 2). It was first published on pages 13–15 of Alexandroff and Urysohn's Mémoire sur les espaces topologique compacts, Verh. Nederl. Akad. Wetensch. Afd. Naturk. Sect. I 14 (1929),1-96.

Its most prominent features are that it is a first countable compact Hausdorff space that contains a discrete subset of cardinality continuum (the circle $C_2$). Thus $X$ can't be second countable and it isn't separable either. Moreover, $C_2$ is open but not a countable union of closed sets, so $X$ is normal but not perfectly normal.

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