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June 2nd, 2014, 01:15 PM   #1
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Open sets in R2

Question: Let a < b, c < d, where all of a, b, c, d are in R. Prove that X = (a, b) x (c, d) is open in R2.

Would it suffice to show that X is open horizontally in (a, b), then vertically in (c, d), then somehow put them together? And if so, how would I go about that?
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June 2nd, 2014, 01:56 PM   #2
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How about this: (x, y) is in X if a < x < b and c < y < d. Construct a neighborhood of (x, y) which is entirely in X.
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June 11th, 2014, 08:50 AM   #3
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Quote:
Originally Posted by nickgarciataria View Post
Question: Let a < b, c < d, where all of a, b, c, d are in R. Prove that X = (a, b) x (c, d) is open in R2.

Would it suffice to show that X is open horizontally in (a, b), then vertically in (c, d), then somehow put them together? And if so, how would I go about that?

the definition of X being open in $\displaystyle R^2$ is the following:

X is is open in $\displaystyle R^2$ iff for all xεΧ there exists r>0 ,such that :

the ball B(x,r) is inside X.

...................................OR............. ........................................

for all ,yεB(x,r) ,yεX.

.....................................OR........... .........................................

For all ,y if ||y-x||<r ,then yεX.............................1

Now since X={(x,y) : xε(a,b) ,yε(c,d)} = {(x,y): a<x<b, c<y<d},if we put :

x=$\displaystyle (x_{1},x_{2})$ and y=$\displaystyle (y_{1},y_{2})$ then the problem is converded to the following problem.

GIVEN , xεX or $\displaystyle a<x_{1}<b , c<x_{2}<d$.................................................. ...............2, we must find an r>o such that :


if $\displaystyle \sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r$,then yεX, or $\displaystyle a<y_{1}<b,c<y_{2}<d$, from (1) and using the definition of the norm in$\displaystyle R^2$

But, $\displaystyle \sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r\Longrightarrow |y_{1}-x_{1}|<r \wedge |y_{2}-x_{2}|<r$, since :

$\displaystyle |y_{1}-x_{1}|<\sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r$

AND:

$\displaystyle |y_{2}-x_{2}|<\sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r$.

BUT:

$\displaystyle |y_{1}-x_{1}|
<r\Longrightarrow x_{1}-r<y_{1}<x_{1}+r$

SO if we want $\displaystyle a<y_{1}<b$ we must have the following relations holding:

$\displaystyle x_{1}+r<b\Longrightarrow r<b-x_{1}$
................................AND............... ............................................

$\displaystyle a<x_{1}-r\Longrightarrow r<x_{1}-a$

In the same way we must have:

$\displaystyle r<d-x_{2}\wedge r<x_{2}-c$

Now we observe that by using (2),

$\displaystyle b-x_{1},x_{1}-a,d-x_{2},x_{2}-c$ are all bigger than zero

Hence if we choose r , 0<r= min{ $\displaystyle b-x_{1},x_{1}-a,d-x_{2},x_{2}-c$}

we are done
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June 19th, 2014, 04:57 PM   #4
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Quote:
Originally Posted by outsos View Post
the definition of X being open in $\displaystyle R^2$ is the following:

X is is open in $\displaystyle R^2$ iff for all xεΧ there exists r>0 ,such that :

the ball B(x,r) is inside X.

...................................OR............. ........................................

for all ,yεB(x,r) ,yεX.

.....................................OR........... .........................................

For all ,y if ||y-x||<r ,then yεX.............................1

Now since X={(x,y) : xε(a,b) ,yε(c,d)} = {(x,y): a<x<b, c<y<d},if we put :

x=$\displaystyle (x_{1},x_{2})$ and y=$\displaystyle (y_{1},y_{2})$ then the problem is converded to the following problem.

GIVEN , xεX or $\displaystyle a<x_{1}<b , c<x_{2}<d$.................................................. ...............2, we must find an r>o such that :


if $\displaystyle \sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r$,then yεX, or $\displaystyle a<y_{1}<b,c<y_{2}<d$, from (1) and using the definition of the norm in$\displaystyle R^2$

But, $\displaystyle \sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r\Longrightarrow |y_{1}-x_{1}|<r \wedge |y_{2}-x_{2}|<r$, since :

$\displaystyle |y_{1}-x_{1}|<\sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r$

AND:

$\displaystyle |y_{2}-x_{2}|<\sqrt{(y_{1}-x_{1})^2+(y_{2}-x_{2})^2}<r$.

BUT:

$\displaystyle |y_{1}-x_{1}|
<r\Longrightarrow x_{1}-r<y_{1}<x_{1}+r$

SO if we want $\displaystyle a<y_{1}<b$ we must have the following relations holding:

$\displaystyle x_{1}+r<b\Longrightarrow r<b-x_{1}$
................................AND............... ............................................

$\displaystyle a<x_{1}-r\Longrightarrow r<x_{1}-a$

In the same way we must have:

$\displaystyle r<d-x_{2}\wedge r<x_{2}-c$

Now we observe that by using (2),

$\displaystyle b-x_{1},x_{1}-a,d-x_{2},x_{2}-c$ are all bigger than zero

Hence if we choose r , 0<r= min{ $\displaystyle b-x_{1},x_{1}-a,d-x_{2},x_{2}-c$}

we are done
This is all very impressive, but what it boils down to is this:

If $(x,y)$ is in the rectangle $(a,b) \times (c,d)$ choose the side that the point $(x,y)$ is closest to (if more than one side is the closest, pick any of them).

Choose $r$ to be less than the distance to the closest side, and the disk of radius $r$ will be entirely contained in the rectangle.

The same logic applies to any "shape"; for any point $x$ of a set with "no boundary", say the set is $A$, we pick:

$r < \inf(\{|y - x|: y \in \partial A\})$

This infimum is just particularly easy to calculate for a rectangle in a rectangular coordinate system.

************

It turns out that the TOPOLOGY (collection of open sets) can be generated by the open disks, OR the open rectangles. We have seen that any open rectangle contains an open disk entirely within the given rectangle. But also: given an open disk of radius $r$, centered at a point $(x_0,y_0)$, the open rectangle:

$\left[x_0 - \dfrac{r}{\sqrt{2}}, x_0 + \dfrac{r}{\sqrt{2}}\right] \times \left[y_0 - \dfrac{r}{\sqrt{2}}, y_0 + \dfrac{r}{\sqrt{2}}\right]$

lies entirely within the open disk.

It therefore matters not if we define "open" to mean every point of a set $A$ contains an open disk contained within $A$, or an open rectangle contained within $A$, we get the same open sets either way.

The advantage to this, is that we can reduce the study of open sets, to the study of cartesian products of open intervals, meaning we can investigate limits and continuity "one variable at a time".

This way, all that work from first-year calculus doesn't go to waste.

In the long run, the difference is WHAT you want to generalize. The "disk topology" generalizes well to any space where you have some notion of a distance function (a positive-definite real function of two variables that satisfies the triangle inequality, such as the norm induced by an inner product). The "rectangle topology" uses intrinsic properties of the real numbers, and is better suited for studying $\Bbb R^n$, or more generally, products of topological spaces where the topology on each factor space is known.

For $\Bbb R^n$, for ANY non-negative integer $n$, the two agree, and which one you use is largely a matter of convenience.
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