2
$\begingroup$

In the book of Analysis on Manifolds by Munkres, at page 202 question 5, it is asked that

Show that if $M$ is a k-manifold without boundary in $\mathbb{R}^m$, and if $N$ is an l-manifold in $\mathbb{R}^n$, then $M\times N$ is a $k+l$ manifold in $\mathbb{R}^{m+n}$.

and I have done the proof of of the case where both $M$ and $N$ are manifolds without boundary (because I have misread the question initially.)

Then I have check the solution manual of the book (found on internet), and it argues that

enter image description here

However, I cannot understand why if both $U$ and $W$ are open in $\mathbb{H}^k$ and $\mathbb{H}^l$, respectively, then why $U\times W$ cannot be open in $\mathbb{H}^{k+l}$. I mean isn't this a direct result from the theory of topological spaces ?

$\endgroup$
6
  • $\begingroup$ in case of a manifold with boundary the coordinate patches may not be open, but homeomorphic to $ \{x\in \mathbb{R}^{k}: x_k \ge 0\}$ which is relatively open in itself, but not in $\mathbb{R}^{k}$ $\endgroup$
    – Thomas
    Aug 3, 2018 at 7:21
  • $\begingroup$ (and the product of such sets will not have the appropriate form for a coordinate patch) $\endgroup$
    – Thomas
    Aug 3, 2018 at 7:23
  • $\begingroup$ @Thomas I want to ask why for your second comment $\endgroup$
    – Our
    Aug 3, 2018 at 7:25
  • $\begingroup$ @AlvinLepik isn't the product of open sets is open in product topology ? $\endgroup$
    – Our
    Aug 3, 2018 at 7:26
  • $\begingroup$ @Thomas I did not understand you first comment $\endgroup$
    – Our
    Aug 3, 2018 at 7:26

1 Answer 1

5
$\begingroup$

For example, $[0, 1)$ is open in $\mathbb{H}$, but $[0, 1) \times [0, 1) \subseteq \mathbb{H}^2$ (the interior of a square together with two of the sides) isn't open.

The reason this doesn't fit with your intuition might be that the topology on $\mathbb{H}^2$ isn't the product topology. $\mathbb{H}^2 \neq \mathbb{H} \times \mathbb{H}$ because the former contains the pair (-1, 1), but the former doesn't (assuming $\mathbb{H}^2 := \{(x, y) \in \mathbb{R}^2 : y \geq 0\}$).

Also, you might be misreading the proof. The problem says "may not be open", but you say "cannot be open". In archaic English, these two are sometimes taken to be equivalent, but (at least in math) this usage is rare these days. The former means that $U \times V$ may or may not be open, not that it must not be open.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .