# Is there an infinite subset of $\Bbb{R}$ not homeomorphic to any of its proper subsets?

Is there an infinite subset of $$\Bbb{R}$$ that is not homeomorphic to any of its proper subsets? Clearly, any finite subset of $$\Bbb{R}$$ is not homeomorphic to any of its proper subsets by mere cardinality. Also, there are infinite topological spaces that also share this property, e.g. $$S^n$$. So the infinitude condition and the subspace of $$\Bbb R$$ condition are crucial.

What I know.

To make it more concise, let's say that $$(X,\tau)$$ is topologically Dedekind-infinite if there is some proper subset $$S\subsetneq X$$ such that, equipped with the restricted topology, is homeomorphic to $$X$$. The assertion "$$X$$ is topologically Dedekind-infinite" will be denoted as $${\sf TInf}(X)$$ and the assertion "$$X$$ is (Dedekind-)infinite as a set" will be denoted as $${\sf Inf}(X)$$.

If it is true that $$\exists X\subseteq \Bbb R:\lnot{\sf TInf}(X)\land{\sf Inf}(X)$$, then $$\exists X\subseteq\Bbb R:\lnot{\sf TInf}(X)$$ satisfying the following conditions:

1. $$X$$ has empty interior (hence, is totally disconnected)
2. $$X$$ is bounded
3. $$X$$ has no isolated points
4. $$X$$ isn't closed
5. $$X$$ isn't countable

The first condition follows from the fact that, if $$X$$ has an interior point, it must contain an open interval $$I\subsetneq X$$. So, if we consider any homeomorphism $$f:\Bbb{R}\to I$$, we have that $$f\vert_{X}:X\to I\subsetneq X$$ is an embedding that would imply $${\sf TInf}(X)$$. The totally disconnected part follows from the fact that, for any $$x_1 in $$X$$, there has to exist some $$x_1 not in $$X$$ since $$X$$ can't contain open intervals.

The second condition follows from the fact that, if $$\lnot{\sf TInf}(X)$$ and $$\Phi:\Bbb R\to(0,1)$$ is a homeomorphism, then $$\lnot{\sf TInf}(\Phi[X])$$ and $$\Phi[X]$$ is bounded.

The third condition follows from two facts. The first one is that $${\sf Inf}(X')\Rightarrow{\sf TInf}(X)$$ where $$X'$$ is the derived set of $$X$$ (the set of isolated points of $$X$$). This is because, if $$\{x_n\}_{n\in\Bbb N}\subseteq X$$ is a set of distinct isolated points, then $$f:X\to X$$ defined as $$x_n\mapsto x_{n+1}$$ (and $$x\mapsto x$$ elsewhere) is a non-surjective embedding $$X\to X$$ implying $${\sf TInf}(X)$$. The other fact is that, if $$x_0\in X$$ is an isolated point, then $${\sf TInf}(X\setminus\{x_0\})\Rightarrow{\sf TInf}(X)$$ since any non-surjective embedding $$f:X\setminus\{x_0\}\to X\setminus\{x_0\}$$ can be extended as a non-surjective embedding $$f^\sharp:X\to X$$ with $$x_0\mapsto x_0$$ and $$f^\sharp\vert_{X\setminus\{x_0\}}=f$$. By contraposition, the $$\lnot {\sf TInf}$$ condition is preserved when removing finitely many isolated and, since there will always be finitely many by the previous fact, it will be preserved when removing all its isolated points.

The fourth condition follows from the fact that, if $$X$$ satisfies the first three properties being nonempty, being closed would also imply that $$X$$ is compact. Thus, by the characterization of the Cantor space, $$X$$ would have to be homeomorphic to the Cantor space which is $${\sf TInf}$$ (it is self-similar).

The fifth condition follows from the fact that, if $$X$$ was countable, then $$X$$ would be a countable metrizable space with no isolated points so, by Sierpiński's topological characterization of $$\Bbb Q$$, it would be homeomorphic to $$\Bbb Q$$ which is $${\sf TInf}$$.

What I tried.

I basically tried constructing such an infinite set following the philosophy of the answers given in this MO question to no avail since I don't know if it is possible to encapsulate the desired condition as a family of $$\mathfrak{c}$$ simple conditions (as Joel David Hamkins did).

• I suspect Scott's On the existence of totally inhomogeneous spaces does this, although the definition isn't quite the same. Commented Jun 29 at 17:07
• Are you asking about a proper subset of ℝ, or of a proper subset of the infinite subset of ℝ ? (That is, what does the word "its" refer to in the title?) Commented Jun 29 at 17:07
• @DanielAsimov I'm asking about an infinite set $X\subseteq \Bbb R$ such that any proper subset (of $X$) $S\subsetneq X$ is not homeomophic to $X$. Commented Jun 29 at 17:19
• @DanielAsimov Wouldn't the latter reading ('not homeomorphic to any of $\mathbb{R}$'s proper subsets') just be trivially false? Commented Jun 29 at 17:29

Yes, there is such a set.

As you suggest near the end of your question, the way to go about building it is to pin down just $$\mathfrak c$$ different conditions that $$X$$ should satisfy in order for it to work, and then to try to build a set $$X$$ satisfying all these conditions via transfinite recursion. The idea of just looking at all subsets of $$X$$ is the wrong approach, because there may be $$>\!\mathfrak c$$ of these.

But remember that if $$D$$ is a dense subset of $$X$$, then every homeomorphism $$X \rightarrow Y$$ is determined already by its restriction to $$D$$. With this in mind, let's construct $$X$$ as follows. Let's decide from the beginning to arrange things so that $$\mathbb Q \subseteq X$$. Then we will look at all continuous functions $$\mathbb Q \rightarrow \mathbb R$$, and build $$X$$, in a $$\mathfrak c$$-stage transfinite recursion, making choices at each stage to ensure that none of these maps extends to a homeomorphism from $$X$$ to a proper subset of $$X$$.

At stage $$\alpha$$ of the recursion, you're given a map $$f_\alpha: \mathbb Q \rightarrow \mathbb R$$, and two disjoint $$\aleph_0+|\alpha|$$-sized sets, $$X_\alpha \supseteq \mathbb Q$$ (points we are committed to putting into $$X$$) and $$Y_\alpha$$ (points we are committed to putting into the complement of $$X$$). If $$f_\alpha$$ already does not extend continuously to $$X_\alpha$$, you're set (and you don't need to do anything at this stage). If $$f_\alpha$$ does continuously extend to $$X_\alpha$$, then because $$X_\alpha$$ and $$Y_\alpha$$ are both smaller than $$\mathfrak c$$, one of two things must happen: either $$f_\alpha$$ extends to an injection on $$\mathbb R \setminus Y_\alpha$$ or it doesn't. If it doesn't, then you can find some $$x,x' \notin Y_\alpha$$ witnessing this non-injectivity, and put them into $$X_\alpha$$. If instead $$f_\alpha$$ does continuously extend to an injection on $$\mathbb R \setminus Y_\alpha$$, then you can find $$x,y \notin X_\alpha \cup Y_\alpha$$ such that the continuous extension of $$f_\alpha$$ to $$x$$ sends $$x$$ to $$y$$; then put $$x$$ into $$X_\alpha$$ and put $$y$$ into $$Y_\alpha$$. Either way, you can, at stage $$\alpha$$, kill the possibility that $$f_\alpha$$ continuously extends to a homeomorphism from $$X$$ to a proper subset of $$X$$. So in the end, the $$X_\alpha$$'s union up to a subset of $$\mathbb R$$ with the property you want.

Observe that this recursion gives you a $$\mathfrak c$$-sized subset of $$\mathbb R$$. Let me point out that the question of whether there is a smaller-than-$$\mathfrak c$$ set with this property is independent of ZFC, and even independent of ZFC$$+\neg$$CH. On the one hand, Martin's Axiom implies that every infinite set of size $$<\mathfrak c$$ is homeomorphic to one of its proper subsets. On the other hand, in the Cohen model, one can take $$X$$ to be any uncountable set of mutually generic Cohen reals (such a set could have size $$\aleph_1$$, even if $$\mathfrak c$$ is much bigger), and this set will have your property.

• I'm confused about the claim “then $f_α$ continuously extends to an injection on $\mathbb{R}\setminus Y_α$”: your assumption is that for each $x \not\in X_α \cup Y_α$ separately, $f_α$ continuously and injectively extends to $x$, but why is it so for all $x$ simultaneously? For “continuous”, I see why being able to extend to each $x$ separately means we can extend to all at once; but for “injective” I don't see the argument. Commented Jun 29 at 18:47
• @Gro-Tsen: Good point, and thanks for the correction. I edited accordingly. Commented Jun 29 at 19:53
• Thanks! Now I'm convinced. Commented Jun 29 at 21:11