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Just a quick question: If you have two sets $A,B \subset \mathbb{R}$ that are open, that is, for every $p \in A$, there exists an $\varepsilon > 0$ such that $B(p;\varepsilon) \subset A$, is the Cartesian Product of these sets also open?

I am trying to think of a proof, but I am stuck rather quickly. So far I have this:

Proof: Let $A$, $B \subset \mathbb{R}$ be open sets. Let $C = A \times B$. Since $A$ is open, there exists an $\varepsilon > 0$ such that for all $p \in A$, $B(p;\varepsilon ) \subset A$. Since $B$ is open, there exists an $\varepsilon ' > 0$ such that for all $q \in B$, $B(q;\varepsilon ') \subset B$.

In other words, we can make a ball around a point in the $x$ or $y$ direction, but can we also make a ball in both at the same time?

Some guidance would be lovely.

Thanks in advance!

Martin Sleziak
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Nigel Overmars
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    The product of the two balls is contained in $A\times B$. Can you show there is a ball around $(p,q)$ that is contained in the product $B(p;\varepsilon)\times B(q;\varepsilon')$? – Daniel Fischer Jan 27 '14 at 10:17
  • @DanielFischer I tried and I can see why it is true. But I can't write it down, could you give another hint? – Nigel Overmars Jan 27 '14 at 10:32
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    Can you bound $\lvert x_i - y_i\rvert$ using $\lVert x-y\rVert$? – Daniel Fischer Jan 27 '14 at 10:35
  • $|x_i - y_i| \leq \sqrt{\sum_{i=1}^2 (x_i - y_i)^2} = | x - y | < \varepsilon$. So in both 'directions' there exists an $\varepsilon > 0$ such that for all $p \in A \times B$, $B(p;\varepsilon) \subset A \times B$? – Nigel Overmars Jan 27 '14 at 10:42
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    Yep. $\lVert (p',q') - (p,q)\rVert < \varepsilon \Rightarrow (\lvert p'-p\rvert < \varepsilon \land \lvert q'-q\rvert < \varepsilon)$. So $B((p,q);\min {\varepsilon,\varepsilon'}) \subset B(p;\varepsilon)\times B(q,\varepsilon')$. Conversely, every ball contains a product of balls, $B(p;\varepsilon/\sqrt{2})\times B(q,\varepsilon/\sqrt{2}) \subset B((p,q),\varepsilon)$. – Daniel Fischer Jan 27 '14 at 10:47
  • Thank you! I understand it now. Make it an answer and I will gladly accept it. – Nigel Overmars Jan 27 '14 at 10:49

1 Answers1

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Yes, the product of two open sets is open. Generally, that is immediate from the definition of the product topology, but in the situation here, the task is to show that the topology induced by the usual Euclidean metric on $\mathbb{R}^d$ (here, $d = 2$) is at least as fine as the product topology (the converse, that every open ball around a point contains the product of open balls around the coordinates, then shows that the two topologies are identical).

If we have $B(p;\varepsilon) \subset A$ and $B(q;\varepsilon') \subset B$, then $B(p;\varepsilon)\times B(q;\varepsilon') \subset A\times B$. Since the coordinate projections don't increase distance,

$$\lvert x_i-y_i\rvert \leqslant \sqrt{\sum_{j=1}^d \lvert x_j - y_j\rvert^2} = \lVert x-y\rVert,$$

it follows that $B((p,q);\min \{\varepsilon,\,\varepsilon'\}) \subset B(p;\varepsilon)\times B(q;\varepsilon')$, and the product of the two open sets $A$ and $B$ is recognised as open.

The converse, that every open ball in the Euclidean metric contains the product of open balls in the coordinate spaces, follows from

$$\lvert x_i-y_i\rvert < \varepsilon/\sqrt{d} \text{ for all } i \Rightarrow \lVert x-y\rVert = \sqrt{\sum_{j=1}^d \lvert x_j-y_j\rvert^2} < \sqrt{\sum_{j=1}^d \varepsilon^2/d} = \varepsilon.$$

Daniel Fischer
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  • Sir, I would like you to elucidate that how'd you get that B((p,q);min{E,E'}) is a subset of B(p,E)XB(q;E'). – Mehul Chaturvedi Jan 10 '18 at 14:08
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    @MehulChaturvedi Let $\eta = \min { \varepsilon, \varepsilon'}$. Then by definition of the balls, $(x,y) \in B((p,q),\eta) \iff \lvert x-p\rvert^2 + \lvert y-q\rvert^2 < \eta^2$. From this we deduce that $\lvert x-p\rvert < \eta$ and $\lvert y-q\rvert < \eta$, so $(x,y) \in B(p,\eta) \times B(q,\eta)$. But $B(p,\eta) \subseteq B(p,\varepsilon)$ and $B(q,\eta) \subseteq B(q,\varepsilon')$, so $(x,y) \in B(p,\varepsilon) \times B(q,\varepsilon')$. – Daniel Fischer Jan 10 '18 at 14:17
  • @DanielFischer does this proof generalize to any dimension for $A \subset \mathbb{R}^n$ and $B \subset \mathbb{R}^m$? – William Oct 02 '22 at 20:29