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Does an uncountable and discrete subspace of the reals exist?

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up vote 27 down vote accepted

Several of the answers given rely on completeness or compactness, but these properties are not needed. For example there is no uncountable discrete subspace of the irrationals. The reals are a second-countable space, so any subspace is also second-countable, which prevents the subspace from having an uncountable discrete subspace.

The proof is as follows. Suppose that $A$ is a discrete space with a countable basis $(B_j : j \in \mathbb{N})$. We want to prove $A$ is countable. Discreteness of $A$ means that each $a \in A$ is in some basic open set that contains $a$ and no other point in $A$. Thus we can make a map $f$ which sends each $a \in A$ to the least $j \in \mathbb{N}$ such that $B_j = \{a\}$. This is an injection from $A$ to $\mathbb{N}$, so $A$ is countable.

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This is interesting. Does the converse hold? I.e., does every space which is not second-countable have an uncountable discrete subspace? – Rasmus Nov 28 '11 at 12:47
I am not sure. I thought I had an example from set-theoretic topology but I no longer think it works. – Carl Mummert Nov 28 '11 at 13:00
@Rasmus, the Sorgenfrey line ( is not second countable, but a modification of Carl's argument shows that it has no uncountable discrete subspace. – JDH Nov 28 '11 at 14:05
can you check if this solution is correct ? it seems too easy to be true , take a minimal set such that it is uncountable and discrete (let it be $A$) , now if we remove an element from $A$ it is still uncountable and discrete,thus contradiction. – Lucyfer Zedd Jun 7 at 7:15

No. Take an uncountable subset of real numbers, then it has to have some $k$ such that there are uncountably many lie in $[k,k+1]$.

Inductively we define smaller and smaller intervals in which there are uncountably many elements. These can be taken as closed, and their intersection will be nonempty by Cantor's theorem.

Therefore every uncountable set of reals has an accumulation point and thus is not discrete.

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I think this has the same problem that I mentioned in Rasmus's answer: a discrete set may have an accumulation point in the ambient space. And closed intervals need not be compact in the relative topology of the subset. – Nate Eldredge Nov 28 '11 at 23:59
@Nate: Last I recall a closed and bounded interval is compact in $\mathbb R$; and a closed subset of a compact space is compact again. – Asaf Karagila Nov 29 '11 at 5:24
Sorry, I was unclear. Call our set $A \subset \mathbb{R}$. To show $A$ is not discrete, it is not sufficient to show it has an an accumulation point in $\mathbb{R}$ (consider $A = \{1/n : n \in \mathbb{N}\}$ which has an accumulation point in $\mathbb{R}$ and is discrete). You have to show that $A$ has an accumulation point in itself. And you cannot adapt this proof by considering decreasing sets of the form $[a_i, b_i] \cap A$, since these need not be compact when $A$ is not closed in $\mathbb{R}$. – Nate Eldredge Nov 30 '11 at 4:49
@Nate: Yes, you are completely right. An even better counterexample to my argument would be $\mathbb R\setminus\{0\}$ and $a_n=-b_n=-n$. I will come up with a nice correction during the day. – Asaf Karagila Nov 30 '11 at 8:55

Edit: The following "proof" is not correct, as pointed out in the comments. Carl Mummert and Asaf Karagila have given arguments which seem correct to me.

The answer in no.

Let $M\subset\mathbb R$ be an uncountable subspace. Then there is $n\in\mathbb N$ such that $M\cap [-n,n]$ is infinite (otherwise $M$ was countable).

Since $[-n,n]$ is compact, the infinite set $M\cap [-n,n]$ has an accumulation point in $[-n,n]$ and hence in $\mathbb R$. Therefore, $M$ cannot be discrete.

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I don't follow. A discrete set can have an accumulation point. Consider $\{1/n : n = 1, 2, \dots\}$. – Nate Eldredge Nov 28 '11 at 14:58
I think you need to show that $M$ has no accumulation point in itself (instead of in $\mathbb R$), which I believe you have not done. What am I missing? =) – Jens Nov 28 '11 at 15:57
@Nate, Jens: Thanks for pointing this out! – Rasmus Nov 28 '11 at 16:24

Assume $X\subseteq\mathbb{R}$ is uncountable and discrete, ie. every point $x\in X$ is isolated. Thus for each point $x\in X$ there is an open neighbourhood $U_x$ such that $U_x\cap X=\{x\}$. Now, using this and the ordering of $\mathbb{R}$ we can find cover of $X$ consisting of open disjoint sets $\{U_x\}_{x\in X}$ st. $U_x\cap X=\{x\}$ for each $x\in X$. Since $\mathbb{R}$ is separable, this collection must be countable, and hence $X$ is countable, which contradicts the assumption.

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It's worth pointing out that the only reason that $X$ has to be separable is because $\mathbb{R}$ was metrizable to begin with. In general, subspaces of separable spaces need not be separable. – Carl Mummert Nov 28 '11 at 13:04
Carl: You're right, of course. An example of a separable space containing nonseparable uncountable subspace is Niemytzki-Moore plane. And the existence of this subspace in this case can be used for example to show that the Niemytzki-Moore plane is not normal (and hence nonmetrizable). In my answer the crucial fact is that open sets from the cover are disjoint in the bigger space $\mathbb{R}$ -- this cannot be obtained in the case of the Niemytzki-Moore plane. – Damian Sobota Nov 28 '11 at 17:03

Every uncountable set of the reals contains a limit point of itself. In fact, if $A$ is uncountable, all but countably many points of $A$ are limit points of $A$.

See, e.g., here

Of course, discrete sets do not contain any of their limit points.

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Let $S$ be a discrete Subset of $\mathbb R$. By definition $S$ is closed. Since $\mathbb R$ is $\sigma$-compact, that is $\mathbb R = \bigcup_{i=1}^{\infty} K_i$ with $K_i$ compact subspaces of $\mathbb R$, we have that $S \cap K_i$ is compact and discrete, hence finite, so $S = \bigcup_{i=1}^{\infty} S \cap K_i$ must be countable.

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Suppose that $A\subseteq \mathbb{R}$ is discrete. Then, for each $a\in A$, there exists an open interval $I_a\subseteq \mathbb{R}$ such that $I_a\cap A=\{a\}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, pick (via the Axiom of Choice, if necessary) a rational number $q_a$ in each $I_a$ such that $q_a\notin I_b$ for each $b\in A\setminus\{a\}$. Then, the mapping $f:A\rightarrow \mathbb{Q}$ defined by $f(a)=q_a$ is injective, implying that $A$ is countable.

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You actually don't need the axiom of choice for this. – Grumpy Parsnip Nov 28 '11 at 15:41
Basically this is the mixture of my argument and Carl Mummert's one. – Damian Sobota Nov 28 '11 at 17:11

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