Properties that are not preserved under homeomorphism

Question

Homeomorphism establishes a very strong relationship between topological spaces. We know that many important properties such as compactness and connectedness are preserved under homeomorphism, and fundamental groups of such spaces are isomorphic.

I wish to gain some more intuition on the possible limits of homeomorphism, so I was wondering if you have any examples of properties that are not preserved under homeomorphism?

Searching SE I found one such example here, which discusses completeness. I think that Vadim's answer was particularly good, since it shows that while the choice of metric is independent of the existence of homeomorphism, the fact that a space $X$ is homeomorphic to a complete metric space means it is metrizable by some complete metric.

So are there any more properties like the above? Or ones that entirely break under homeomorphism?

Answer 1

What about a sequence being a Cauchy sequence? The spaces $\mathbb R$ and $(0,+\infty)$ are homeomorphic (with respect to the usual metric). A homeomorphism between them is $\exp\colon\mathbb{R}\longrightarrow(0,+\infty)$, whose inverse is $\log$. The sequence $(1/n)_{n\in\mathbb N}$ is a Cauchy sequence in $(0,+\infty)$, but $\bigl(\log(1/n)\bigr)_{n\in\mathbb N}$ is not.

Answer 2

A general homeomorphism does not preserve distances. If we have two metric spaces $X, Y$ and a homeomorphism $f:X\to Y$, $f$ need not be an isometry. For instance, we can take $f:[0,1]\to [0,2]$ given by $f(x)=2x$. This map is not distance preserving, because $$d(1,0)\ne d(f(1),f(0))=d(2,0).$$ The upshot is that properties that rely on a stronger structure than the topology on the space need not be preserved under isomorphism. For instance, the metric structure need not be preserved.

Answer 3

You may study $Isometry$ which is stronger than $homomorphism$,It is a bijective map $f:(X,d_1) \longrightarrow (X,d_2)$ defined as $d_1(a,b)=d_2(f(a),f(b))$;As it preserves the distance between these two spaces it will help you.

Answer 4

You could look at properties describing the differential topology of a differential (as opposed to merely topological) manifold. Surely some properties preserved by diffeomorphisms are not in general preserved by homeomorphisms.

Cf. this question Properties preserved by diffeomorphisms but not by homeomorphisms

Answer 5

(I). A non-compact metrizable space $X$ has an unbounded metric $d,$ and the metric $e(x,y)=\min (1,d(x,y))$ generates the same topology, so $id_X$ is a homeomrphism from $(X,e)$ to $(X,d).$ So being a bounded subset of $X$ is not a topological property: $X$ itself is bounded with respect to $e$ but not with respect to $d.$

(II). For a metric space $(X,d)$ and $Y\subset X$ and $r>0$ let $Y$ be $r$-dense in $X$ iff $\forall x\in X \;\exists y\in Y\;(d(x,y)<r).$ Equivalently $Y$ is $r$-dense in $X$ iff $\cup_{y \in Y}B_d(y,r)=X.$

The function $f(x)=\tan x$ is a homeomorphism from $(-\pi /2,\pi /2)$ to $\mathbb R$ (with the usual metric on both spaces). Any non-empty subset of $(-\pi /2,\pi /2)$ is $\pi$-dense in $(-\pi /2,\pi /2)$ but $\{2^n:n\in \mathbb N\}$ is not $r$-dense in $\mathbb R$ for any $r>0.$

(III).(Digression). When $X$ is compact and metrizable there are some important properties shared by all metrics for $X$: Any metric for $X$ is bounded,complete, totally bounded, and satisfies the Lebesgue covering theorem.

A non-compact metrizable space has an incomplete metric. This is not an easy result.

Answer 6

You may be also interested in other kind of mappings (continuous, quotient, open, closed, opened and closed, perfect, ...) and operations (taking a subspace, or the finite, countable, infinite cartesian product or sum) that preserve topological and topologically-related properties (being $T_0$, $T_1$, ..., locally, sequentially, countably, weakly, strongly (para-) compactness, being lindelof, metrizable, ...).

In such a case you should consult Engelking's General Topology that gathers and presents these questions in outline form at the end of the book.

Answer 7

We have the completeness as well which is not preserve $\Bbb R$ is homeomorphic to $(-1,1)$ which is not complete but $\Bbb R $ is complete. A bijection is given by

$$\phi(x) = \frac{x}{1+|x|} \in (-1,1),~~~~\forall x\in \Bbb R.$$

Answer 8

Two subspaces of a topological space can be homeomorphic while the first is a closed set of the topological space and the other is not.

For example, in $\mathbb{R}^2$ with the usual product topology, $\mathbb{R}$$\times$$0$ $\simeq$ $\mathbb{R}$ $\simeq$ $\mathbb{(0,1)}$ $\simeq$ $\mathbb{(0,1)}$$\times$$0$, where $\simeq$ means homeomorhpism. $\mathbb{R}$$\times$$0$ is closed in $\mathbb{R}^2$ and $\mathbb{(0,1)}$$\times$$0$ is not closed.

Thus, it is possible that two homeomorhpic spaces can be embedded in a topological space, where the first is a closed set there and the other is not.