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I've got a homework to do, I'm not sure if I'm missing parenthesis or something, I've got to prove this:

For the metric spaces $E=(E,d_1)$ and $F=(F,d_2)$ with $A\subset E$ and $B\subset F$.

$$\partial(A\times B)= \partial A\times \overline{B} \cup \overline{A} \times \partial B$$

Please don't let the spaces guide you becasue they mean nothing in special, just latex doing his thing. My best guess is that, taking in account "dimension" it must be the union of the products.

like this $$\partial(A\times B)= (\partial A\times \overline{B}) \cup (\overline{A} \times \partial B)$$

Question is: Am I correct? How would you add parenthesis to make this statement true? (ofc for non trivial sets) Is there any program I can do Venn diagrams for matematicians, like adding closure and interior of sets ?

Andrés E. Caicedo
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José Osorio
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  • I'm pretty sure we can assume the text assumes cross products take precedence over unions so $A \times B \cup C \times D$ by definition means $(A \times B) \cup (C \times D)$. After all, if $A,C \subset X$ and $B,D \subset Y$ thenn $(A \times B) \cup (C \times D) \subset X \times Y$ whereas $A \times (B \cup C) \times D$ most certainly is not. – fleablood Apr 16 '16 at 16:47
  • Thank you, about the other question, know anything about a software to study sets?, I can't seem to find a realiable list of algebra of sets for closure, boundaries and interior points. – José Osorio Apr 16 '16 at 16:54
  • I don't know anything about math software. I think I can give you some advice on how to prove this with just analysis and definitions if you need it. Do you? – fleablood Apr 16 '16 at 18:06
  • I've got my own try but yes I'd appreciate the help, theres a moment in the definition of closure of set in which you have to use the union of all the closed sets that cointain $A\times B$, but I don't know how to make it "talk" about the "coordinates" only, and not he "vector" x. (sorry for using comillas, my english vocabulary is small heh) – José Osorio Apr 16 '16 at 18:49
  • Actually we need some basic idea what metric, $d_3$ we are going to use on $X \times Y$. I think the exercise is assume that whatever metric we chose, if restricted to a "plane", it will be compatible with $d_1$ and $d_2$. In other words $d_3((x,y),(x,z)) = d_2(y,z)$ and $d_3((x,y),(z,y)) = d_1(x,z)$. I think if $d_3$ doesn't obey this property we can't conclude anything, but I 'm not going to waste any brain cells coming up with a counter-example. $d_3 = \sqrt {d_1^2 + d_2^2}$ and $d_3 = d_1 + d_2$ are both possible candidates. As long as $d_3$ has that property we can prove this...tbc.. – fleablood Apr 16 '16 at 20:13
  • what! you mean i can't do this by simply using set theory? – José Osorio Apr 16 '16 at 20:22
  • If $d_3$ has the property $d_3((x,y)(x,z)) = d_2(y,z)$ and $d_3((x,y)(z,y)) = d_1(x,z)$ then it's easy to prove if B is a neigbhorhood of (x,y) then B restricted to {x} $\times Y$ then $B \cup {x} \times Y = {x} \times B'$ where $B'$ is a neigborhood of y in Y . Similarly for B restricted to X $\times$ {y}. From there I think the prove is simply taking cases. Prove if x in d(A) and y in clo(B) then as every neighborbood of x contains point is A and comp (A) then it contains points of comp (A x B) and points of A x {y} and as y is in clo(B) if y in B the nbrhd contains points of AxB. ... – fleablood Apr 16 '16 at 20:24
  • Well, without a metric or a topology on X x Y the question is meaningless. I'm assuming the exercise is assuming the metric and/or topology is compatible with the metric of X and Y when restricted to X and Y. If so then, yes, you certainly can do this with just set theory. If not, you can't do this at all. – fleablood Apr 16 '16 at 20:27

1 Answers1

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I am assuming that there is a metric, $d_3$ on $X \times Y$ that when restricted to a point cross X is $d_1$ and when restricted to a point cross Y is $d_2$. i.e $d_3((x,y),(x,z)) = d_2(y,z)$ and $d_3((x,y),(z,y)) = d_1(x,z)$.

Claim: if $B(x,y)_{\epsilon}$ is the neighborhood of all {$(w,z)\in X \times Y|d_3((x,y),(w,z) < \epsilon$} and if $(a,b) \in B(x,y)_{\epsilon}$. Then there exist $\epsilon' > 0$ and $\epsilon" > 0$ such that $B(a)_{\epsilon'} = \{w \in X| d_1(a,w) < \epsilon'\}$ and $B(a)_{\epsilon'} \times \{b\} \subset B(x,y)_{\epsilon}$ and $B(b)_{\epsilon"} = \{z \in Y| d_2(b,z) < \epsilon"\}$ and $\{a\} \times B(b)_{\epsilon"} \subset B(x,y)_{\epsilon}$

Pf: Let $\epsilon' = \epsilon - d_3((x,y),(a,b))$. If $d_1(a,w) < \epsilon'$ so $a \in B(a)_{\epsilon}$ and $(a,b) \in B(a)_{\epsilon'} \times \{b\}$, then $d_3((w,b),(x,y)) \le d_3((w,b),(a,b)) + d_3((a,b),(x,y)) = d_1(a,w) + d_3((a,b),(x,y)) < \epsilon' + d_3((a,b),(x,y)) = \epsilon$.

So $B(a)_{\epsilon'} \times \{b\} \subset B(x,y)_{\epsilon}$.

Proving $B(b)_{\epsilon"} = \{z \in Y| d_2(b,z) < \epsilon"\}$ is exactly the same.

So let's start.

Let $(x,y) \in \partial A \times \overline B$ which is the same thing as stating $x \in \partial A$ and $y \in \overline B$. Let $N$ be a neighborhood of $(x,y)$ and $N'(x)$ a neighborhood of $x$ so that $N'(x) \times \{y\} \subset N$ and $N"(y)$ a neighborhood of $y$ so that $\{x\} \times N"(y) \subset N$.

As $x \in \partial A$ there are points $w \in N'(x)$ such that $w \not \in A$ so $(w,y) \in N$ but $(w,y) \not \in A \times B$. So $(x,y)$ is a limit point of $(A \times B)^c$.

Now $x \in \partial A$ so there are points $v \in N'(x)$ such that $v \in A$. If $y \in B$ then $(v,y) \in A \times B$. If $y \not \in B$ then $y$ is a limit point of $B$. Let $\delta < |radius(N') - radius(N")| < radius(N")$. Then there is a $z \in B$ such that $d_2(y,z) < \delta$ so $(v,z) \in A \times B$ and $v \in N'(x)$ and $z \in N"(y)$ so $(x,z) \in N$. So $(x,y)$ is a limit point of $A \times B$.

So $(x,y) \in \partial(A\times B)$ and $\partial A \times \overline B \subset \partial(A\times B)$.

Similaraly $\overline A \times \partial B \subset \partial(A\times B)$. So $\partial A \times \overline B \cup \partial A \times \overline B \subset \partial(A\times B)$.

That's half of it...

fleablood
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  • Wow it's just ... i have now idea how you had the idea to do this, i was barely thinkin in each definition and working with that, to use B "balls" in this is beyond intuition for me... can you give me a head start for the other part? – José Osorio Apr 19 '16 at 00:59