Are they topological invariants?

Question

In my topology course, my teacher left us a list of property and asked us to prove or disprove that they are actually topological invariants, i.e. properties that would remains invariants under a homeomorphism, say $f: X \rightarrow Y$.

1. Openess of $S \subset X$
2. Closedness of $S \subset X$
3. Compactness of $X$
4. Completeness of $X$
5. Boundedness of $S \subset X$
6. Distance, i.e. $d(x, y)$ for all $x,y \in X$

I have some thoughts in mind: (1), (2), (3) are topological invariants, (6) is not, while I don't know the others

(1)&(2): Note that $f$ being a homeomorphism, both $f$ and $f^{-1}$ should be continuous. Also note that under continuous function, open(closed) images have open(closed) pre-images. It quickly implies that openess and closedness are preserved under homeomorphism.

(3): It is true with a similar argument as above. (Compactness should be preserved under continuous function.)

(6): Consider $f: \{(x,y):x=y\}\rightarrow \mathbb{R}$, $(x,x) \mapsto x$, which is a homeomorphism. $d((1,1),(2,2)) = \sqrt 2 \neq 2-1$.

In fact, I haven't seen many examples of homeomorphisms. It would be great if you can tell me if I am correct in the above and guide me through (4) and (5). Thanks in advance.

Answer 1

(5) is not a topological invariant. It depends on the metric. In fact, if $(X,d)$ is a metric space then $d'(x,y)=d(x,y)/[1+d(x,y)]$ is a metric on $X$ and $(X,d)$ and $(X,d')$ are homeomorphic. Especially, every subset of $X$ is bounded under the metric $d'$ but $X$ can have unbounded subset under the metric $d$.

(4) is also not a topological invariant. It is known that the set of irrationals under normal metric (the metric given by well-known absolute value) and Baire space is homeomorphic. But first one is not complete and second one is complete. (Note that Baire space has a metric defined as $$d(\langle x_n\rangle_{n\in \Bbb{N}},\langle y_n\rangle_{n\in \Bbb{N}})=2^{-\min\{k\in\Bbb{N} : x_k\neq y_k\}}.$$

Answer 2

(4) $\mathbb R$ is complete and $(0,1)$ is not (both with usual metric). But the two spaces are homeomorphic!

This also serves as counterexample to (5) and (6)

Answer 3

An alternative conterexample for (4) is $f(x)=x/(x^2-1)$ for $x\in(-1,1)$. It easy to check that $f:(-1,1)\to\Bbb R$ is a homeomorphism, and the subspace $(-1,1)$ is not complete, but $\Bbb R$ is.

Answer 4

Another example for completeness: let $X=\Bbb Z^+$ and $Y=\left\{\frac1n:n\in\Bbb Z^+\right\}$, both with the usual metric inherited from $\Bbb R$.

• Show that $X$ is complete, because a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ is Cauchy if and only if it is eventually constant, i.e., if and only if there is an $m\in\Bbb N$ such that $x_n=x_m$ for all $n\ge m$.

• Show that $Y$ is not complete.

• Show that the map $h:X\to Y:n\mapsto\frac1n$ is a homeomorphism. (If you’re more ambitious, you can show that every bijection between $X$ and $Y$ is a homeomorphism.)

This of course also gives you an example for boundedness.