# Non-homeomorphic spaces that have continuous bijections between them

What are nice examples of topological spaces $X$ and $Y$ such that $X$ and $Y$ are not homeomorphic but there do exist continuous bijections $f: X \to Y$ and $g: Y \to X$?

Recycling an old (ca. 1998) sci.math post:

" Anyone know an example of two topological spaces $X$ and $Y$ with continuous bijections $f:X\to Y$ and $g:Y\to X$ such that $f$ and $g$ are not homeomorphisms?

Let $X = Y = Z \times \{0,1\}$ as sets, where $Z$ is the set of integers. We declare that the following subsets of $X$ are open for each $n>0$. $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a topology on $X$.

We declare that the following subsets of $Y$ are open for each $n>0$. $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a toplogy on $Y$.

Define $f:X\to Y$ and $g:Y\to X$ by $f((n,i))=(n,i)$ and $g((n,i))=(n+1,i).$ Then $f$ and $g$ are continuous bijections, but $X$ and $Y$ are not homeomorphic.

This example is due to G. Paseman.

More generally, take a space X with three successively finer topologies T, T' and T''. Form two spaces which have underlying set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' .... and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer topology in one sequence to a rougher topology in the other. You can make them bijective, and show that they are obviously non-homeomorphic for a judicious choice of X, T, T', and T''.

• If you want non-compact variations on this, e.g. making the two big spaces normal or connected, then choose X and the T's even more judiciously, and add an extra point or two as needed. And check your work, of course! I leave the compact version(s), or their impossibility, to you. Gerhard "Ask Me About System Design" Paseman, 2010.07.05 Commented Jul 5, 2010 at 23:19

I know this is super old, but somebody asked the same question again (Non-homeomorphic topological spaces) and so I wanted to share a "proof by picture" that settles the question.

(I came up with this a few years ago when I was grading an exam in which somebody claimed that $X$ and $Y$ are necessarily homeomorphic.)

Of course, this is just a special case of the accepted answer above. But I think it's nice to picture it in a geometric way...

• This pictoric view looks like the Eilenberg swindle. Commented Sep 29, 2016 at 18:26
• all you really need to show is the last 3 figures. Commented Jan 28, 2017 at 5:37
• Aren’t $X$ and $Y$ homeomorphic? They’re both infinite products of $S^1$ and $[0,1)$. Commented Nov 25, 2021 at 13:57
• @D.Wynter: They're not infinite products, they're infinite disjoint unions of some specific finite products of those. Commented Dec 9, 2021 at 3:47

Here's a continuum analogue of Gerhard Paseman's answer: Let $X$ and $Y$ be topological spaces whose underlying sets are $\mathbb{R}$. As topological spaces, $X$ is the disjoint union of the open interval $(0,\infty)$ with a discrete space whose points are nonpositive reals, while $Y$ is the disjoint union of $(-1,0)$, $(1,\infty)$, and a discrete space whose points form the complement of those intervals. Translation by adding one is a continuous bijection from $X$ to $Y$, and also a continuous bijection from $Y$ to $X$, but the two spaces are not homeomorphic.

• I like the example, but I do not see a good proof that they are not homeomorphic. Is e.g. (3,5) an open set in both X and Y, or just a subset of an open set? Gerhard "Ask Me About System Design" Paseman, 2010.07.05 Commented Jul 6, 2010 at 6:02
• Look at the one-point compactifications, after noting that both spaces are locally compact Hausdorff. For $X$ we get a disjoint union of a circle and uncountably many points, while for $Y$ we get a disjoint union of a wedge of two circle and uncountably many points. If $X$ and $Y$ were homeomorphic, their one-point compactifications would be as well. Commented Jul 6, 2010 at 9:13
• Look at the number of non-singleton connected components: $X$ has one and $Y$ has 2. This is a topological invariant. Commented Jul 7, 2010 at 11:08

I've asked myself this question some time ago and found some counterexamples. After having done this I asked myself what the "smallest" counter example would be. First of all notice that if the set of opens in $X$ and in $Y$ have to have the same cardinality. This show that that if the number of opens in $X$ is finite then $f$ and $g$ both have to be homeomorphisms so the cardinality of the set of opens needs to be at least $\aleph_0$. The answer of Gerhard Paseman shows that using $2^{\aleph_0}$ opens is enough. But one can do better, here is an example that actually has $\aleph_0$ opens:

Let $X$ and $Y$ both have $\mathbb Z$ as underlying set, let the opens of $X$ be $\lbrace \mathbb Z_{\geq i} \mid i \in \mathbb N_{>0} \rbrace \cup \lbrace \emptyset, \mathbb Z \rbrace$ and the set of opens of $Y$ be $\lbrace \mathbb Z_{\geq i} \mid i \in \mathbb N_{>0} , i\neq 2\rbrace \cup \lbrace \emptyset, \mathbb Z \rbrace$, then $f:X\to Y$ given by $f(x)=x$ is continues, and so is $g:Y\to X$ given by $g(a)=a-2$. To see that $X$ and $Y$ are not homeomorphic note that $X$ contains only 1 element that is contained in exactly 2 opens, namely the element $1$, but in $Y$ both the element $1$ and the element $2$ are contained in exactly 2 opens.

You may find the paper "Bijectively related spaces. I. Manifolds" by P. H. Doyle and J. G. Hocking intersting. They cite some related work, which is also worth checking.

There is a later paper "Unusual and bijectively related manifolds" by J. G. Hocking on this topic, which I have not read.

If you are also interested in spaces satisfying only low separation axioms, you may try my own humble work here and here.

I just posted an answer to this related MO question. To sum up the part that's relevant here:

Let $\mathcal N$ be the Baire space and let $X$ be any other zero-dimensional Polish space that is not $\sigma$-compact. Then there are continuous bijections $\mathcal N \to X$ and $X \to \mathcal N$.

I don't have a short proof of this assertion, but you can look at my paper for a proof, and at another paper of mine (joint with Arnie Miller) where some similar things are explored.

Now, let's fix a particularly nice $X$: say $X$ is the disjoint sum of $\mathcal N$ and the Cantor space $\mathcal C$. For this special case, I can give you a short proof that they're bijectively related (modulo a few well-known results).

To get a continuous bijection $\mathcal N \to X$:

By Exercise 7.15 in Kechris's Classical Descriptive Set Theory, a nonempty Polish space $Y$ is perfect if and only if there is a continuous bijection $\mathcal N \to Y$. $X$ meets these requirements.

To get a continuous bijection $X \to \mathcal N$:

There is a homeomorphic copy of $\mathcal C$, say $K$, with $K \subseteq \mathcal N$. By Theorem 7.7 in Kechris's book, $\mathcal N \setminus K$ is homeomorphic to $\mathcal N$. Thus we can get a continuous bijection $X \to \mathcal N$ by mapping $\mathcal C$ homeomorphically onto $K$ and mapping $\mathcal N$ homeomorphically onto $\mathcal N \setminus K$.

I don't have my copy of Kelley handy but I think in chapter 1 he gives the example where X is a countable disjoint union of open intervals and a countable discrete set while Y is a countable disjoint union of left-closed, right-open intervals and a countable discrete set.The point is that you can get a half closed interval from an open one by attaching an endpoint and you can build an open interval using a sequence of half closed intervals.

Here is an example which comes from using the spaces of Charles Siegel's post.

One have a continuous bijection from [0,1) to the circle given by the exponential function ( t-->exp(2ipit) ). The idea is to use this to construct our spaces. Take A to be a wedge of countably many (one for each integer) [0,1) attached at 0. Let A_n be the same wedge but replacing the [0,1) corresponding to the integers from 1 to n by circles.

X is going to be the disjoint union of A_2, A_4, A_6,... and countably many copies of A. Y is going to be the disjoint union of A_1, A_3, A_5,... and countably many copies of A.

We have a continuous bijection from A_n to A_{n+1} given by replacing the copy of [0,1) corresponding to the integer n+1 by a circle (as it is in A_{n+1}). Using this maps we are going to define f and g.

f is going to map one copy of A to A_1, A_2 to A_3, A_4 to A_5,... and so on, and the rest of the copies of A to the rest of the copies of A in Y. g is going to map A_1 to A_2, A_3 to A_4,... and so on, and the copies of A to the copies of A.

What about $X= \mathbb R$ with usual topology and $Y=[a,b]$ with subspace topology? Then $(a,b)$ is a subspace of $[a,b]$ which is homeomorphic to $\mathbb R$ and $[a,b]$ is a subspace of $\mathbb R$ which is homeomorphic to $[a,b]$.

My favorite, which is on the wikipedia page for "homeomorphism", is $\phi:[0,2\pi)\to S^1$, by $\phi(\theta)=(\cos\theta,\sin\theta)$, which is continuous and bijective, but not a homeomorphism.

• The OP also wants a continuous bijection in the other direction. Commented Jul 5, 2010 at 22:38
• Ahh! Missed that. My apologies. Commented Jul 5, 2010 at 22:40