Understanding Convex Open Sets in Multivariable Calculus

[geoffrey159]

I was trying to understand a proof in multivariable calculus (about fundamental pde's), and at some point it said that if $U$ is a convex open set of $\mathbb{R}^2$, then it's second projection is an open set in $\mathbb{R}$, without any further explanation. Could you explain please ?

 

[MarcusAgrippa]

I was trying to understand a proof in multivariable calculus (about fundamental pde's), and at some point it said that if $U$ is a convex open set of $\mathbb{R}^2$, then it's second projection is an open set in $\mathbb{R}$, without any further explanation. Could you explain please ?

Open means that the set includes all points in its interior, and does not include any of its boundary points.
Convex means that, if you join any two points in the set with a line segment, the line segment lies completely in the interior of the set.

There are more precise definitions of both concepts, but intuitively the ideas are as stated above.

 

[geoffrey159]

Think I've got more on this :

For any point $a = (x_0,y_0) \in U$ there is an open ball $B(a,r)$ included in $U$.
For any point $u = (x,y)$ in the ball, its symetric point by $a$, called $v= (2x_0 -x, 2y_0 -y )$ is also in the ball, and by convexity of the ball, the lign joining $u$ and $v$ is also in the ball.
This lign is the set $S = \{ 2(1-\lambda)a + (2\lambda -1) u,\ \lambda\in [0,1] \}$.
So it appears to me that $\text{pr}_2(S)=\{ 2(1-\lambda)y_0 + (2\lambda -1) y,\ \lambda\in [0,1] \}$ is a segment containing $y_0$ as its middle point. So clearly, there is an open ball centered in $y_0$ included in $\text{pr}_2(U)$, which proves this set is open in $\mathbb{R}$.

Does it work for you ?

 

[WWGD]

This is a technical issue in part. The open sets in $\mathbb R^2$ (assumed product topology) have the form $\cup U_i \times V_i$ , for $U_i,V_i$ open in $\mathbb R$. Then it is to show that the projection of any set of this type is open in $\mathbb R$. A path in $\mathbb R^2$ between $x=(a_1,b_1),y=( a_2,b_2)$ is a continuous map , usually from $[a,b] \rightarrow R^2 =(f_1(x),f_2(x)) (f_1(0)=a_1, f_2(0)=b_1 , f_1(1)=a_2 ,f_2(1)=b_2$ show that if a path between $x,y$ lies entirely inside of the open set, then its projection lies entirely in the projected open set.

 

[geoffrey159]

Thank you, but why do you assume that $U$ has the form of a cartesian product of open sets ?
And also, I doubt that all open sets in $\mathbb{R}^2$ have the form $\cup \ U_i\times V_i$, $U_i,V_i$ open sets in $\mathbb{R}$ ( I am not convinced that $B(0,1)$ be described in such a way, can it ? )

 

[pasmith]

Thank you, but why do you assume that $U$ has the form of a cartesian product of open sets ?

Because this is the definition of the product topology.

And also, I doubt that all open sets in $\mathbb{R}^2$ have the form $\bigcup \ U_i\times V_i$, $U_i,V_i$ open sets in $\mathbb{R}$ ( I am not convinced that $B(0,1)$ be described in such a way, can it ? )

For every (x,y) \in B(0,1), there is an open rectangle (x - \delta, x + \delta) \times (y - \epsilon, y + \epsilon) centered at (x,y) and contained entirely within B(0,1). Take the union of these rectangles: it contains all of B(0,1) (every point of B(0,1) lies within at least one rectangle) and nothing but B(0,1) (each of these rectangles is contained within B(0,1)).

 

[geoffrey159]

Hmmm, this looks suspicious :-)
But since I don't know a great deal about topology I'm not going to insist. However, with minimal knowledge as developped in post #3, can I conclude ?

 

[WWGD]

Thank you, but why do you assume that $U$ has the form of a cartesian product of open sets ?
And also, I doubt that all open sets in $\mathbb{R}^2$ have the form $\cup \ U_i\times V_i$, $U_i,V_i$ open sets in $\mathbb{R}$ ( I am not convinced that $B(0,1)$ be described in such a way, can it ? )

No, I said every open set is _the union_ of product sets, not necessarily ( as you correctly pointed out) a product itself. Still, any interior point is contained in a "product ball " (a square, actually; open set "can spare a square") $U \times V$ inside of the open set. I think an argument by contradiction: assume the open set projects into a non-open set $W$, containing some non-interior point $y$ . What can you say about $\pi_2^{-1}(y)$? And in $\mathbb R$, every open ball is convex -- no room to step out of the open ball in the Real line.

 

[mathwonk]

to detail what has already been said:

1. the projection from R^2 to R of any open set U in R^2 is open in R. proof: if (a,b) lies in the open set U, it suffices to show that b lies in the interior of the projection of U. by definition of the product topology, there is a recrtangle of form AxB that lies entirely in U, where A and B are open intervals in R, and a lies in A, and b lies in B. Then the projection of U contains the projection of AxB, namely B, which is an interval of R containing b. QED.

2. Since the proof above does not use the property of convexity of U, I suspect that the statement you want is that the projection of U is a convex open set in R, namely an interval. This has been essentially argued in the previous posts as well. This follows from the fact that the projection of a line segment contained in U, is a line segment contained in the projection of U.

 

[geoffrey159]

No, I said every open set is _the union_ of product sets, not necessarily ( as you correctly pointed out) a product itself. Still, any interior point is contained in a "product ball " (a square, actually; open set "can spare a square") $U \times V$ inside of the open set. I think an argument by contradiction: assume the open set projects into a non-open set $W$, containing some non-interior point $y$ . What can you say about $\pi_2^{-1}(y)$? And in $\mathbb R$, every open ball is convex -- no room to step out of the open ball in the Real line.

Your remark about the nature of open sets in $\mathbb{R}^2$ makes the proof even simpler :
Assuming that $U = \cup_{i \in I}\ U_i \times V_i$, $U_i$ and $V_i$ all open in $\mathbb{R}$, then for any pair $(x,y) \in U$, there exists $i\in I$ such that $y$ belongs to $V_i$.
Therefore, there exists a subset $I_0\subset I$ such that $\pi_2(U)\subset \cup_{i\in I_0} V_i$.
Conversely, for any $y\in \cup_{i\in I_0} V_i$, there exists $i\in I_0$ such that $y\in V_i$, and since $U_i \times V_i \subset U$, it is clear that there exists $x\in U_i$ such that $y=\pi_2(x,y) \in\pi_2(U)$. Therefore $\cup_{i\in I_0} V_i \subset \pi_2(U)$.
Finally, $\pi_2(U) = \cup_{i\in I_0} V_i$ which is open as a union of open sets.

 

[WWGD]

Your remark about the nature of open sets in $\mathbb{R}^2$ makes the proof even simpler :
Assuming that $U = \cup_{i \in I}\ U_i \times V_i$, $U_i$ and $V_i$ all open in $\mathbb{R}$, then for any pair $(x,y) \in U$, there exists $i\in I$ such that $y$ belongs to $V_i$.
Therefore, there exists a subset $I_0\subset I$ such that $\pi_2(U)\subset \cup_{i\in I_0} V_i$.
Conversely, for any $y\in \cup_{i\in I_0} V_i$, there exists $i\in I_0$ such that $y\in V_i$, and since $U_i \times V_i \subset U$, it is clear that there exists $x\in U_i$ such that $y=\pi_2(x,y) \in\pi_2(U)$. Therefore $\cup_{i\in I_0} V_i \subset \pi_2(U)$.
Finally, $\pi_2(U) = \cup_{i\in I_0} V_i$ which is open as a union of open sets.

Good going, nice and clear, you got it.

 

[geoffrey159]

Thank you ! This was very informative, it deserved some explanations