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It is a basic fact of topology that if $X$ is a topological space and $Y\subseteq X$ is homeomorphic to $X$, it does not need to occur that $X=Y$ (for example, $X=\mathbb{R}$, $Y=(0,1)$). My question is, if I add the requirement that $Y$ is dense in $X$. Is this still the case? Or is the following true?

If $X$ is homeomorphic to a dense subspace $Y\subseteq X$, then $X=Y$.

Saúl Pilatowsky-Cameo
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4 Answers4

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It's not true. Consider $Y=\mathbb N$ and $X=\mathbb Z$, both with the trivial (indiscrete) topology.

Then $Y$ is dense in $X$ (every non-empty subset is dense in this topology) and homeomorphic to $X$

(there is a bijection between $Y$ and $X$, and every map to an indiscrete space is continuous),

but clearly $X\ne Y$.

J. W. Tanner
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For a more interesting example, let $X=\Bbb Q$ with the usual topology, and let

$$Y=\left\{\frac{m}{2^n}\in\Bbb Q:m,n\in\Bbb Z\text{ and }n\ge 0\right\}$$

be the set of dyadic rationals. Then $Y$ is a proper subset of $X$ that is both dense in $X$ and homeomorphic to $X$.

Brian M. Scott
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  • How do you show $Y$ is homeomorphic to $X$? Is every dense subset of $\mathbb{Q}$ homeomorphic to $\mathbb{Q}$? – Jair Taylor Nov 18 '20 at 06:14
  • @JairTaylor: There are several ways. For instance, it’s a theorem that the rationals are the unique countable metric space without isolated points. Alternatively, every countable dense linear order without endpoints is order-isomorphic to $\Bbb Q$, and the subspace topology on $Y$ is also its natural order topology. – Brian M. Scott Nov 18 '20 at 06:20
  • I see, thank you! – Jair Taylor Nov 18 '20 at 07:51
  • @JairTaylor: You’re welcome! – Brian M. Scott Nov 18 '20 at 16:00
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Counterexample:

$$\mathbb{Q}\backslash \{0\} \subset \mathbb{Q}$$

$\bf{Added:}$ It turns out that if $A$, $B$ are dense, countable subsets of $\mathbb{R}$, then there exists a homeomorphism $f\colon \mathbb{R}\to \mathbb{R}$, such that $f(A)=B$.

Perhaps an explicit example is best. Consider $A=\mathbb{Q}$, and $B = \mathbb{Q}^3\subset \mathbb{Q}$. The map $x\mapsto x^3$ is a homeomorphism of $\mathbb{R}$ taking $A$ to $B$.

J. W. Tanner
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orangeskid
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Let $X=\mathbb{Z}$ and $Y=\mathbb{N}$ then $\mathbb{N}\subset \mathbb{Z} $, and define trivial topology $T$=$\{\emptyset,\mathbb{Z} \}$ on $\mathbb{Z}$. Let $x \in \mathbb{Z}$ , then $\forall$ open neighbourhood of $x$ there is non empty intersection with $\mathbb{N}$. Therefore $\mathbb{N}$ dense in $\mathbb{Z}$. Also let define function $f :\mathbb{N}$ $\rightarrow$ $\mathbb{Z}$ as follows

$f(n) =\begin{cases} -(n-2)/2, & \text{if $n$ is even} \\ (n+1)/2, & \text{if $n$ is odd} \end{cases}$

Therefore $f$ is homeomorphic function. So in here clearly $\mathbb{Z}$ $\neq$ $\mathbb{N}$

Lahiru Dhananjaya
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