# How can I use the concept of a proper map to show that a set is bounded?

• I
• cbarker1
In summary: You can see this by noting that they are equivalent to x^2 = 1 or y^2 = 1, respectively. On the other hand, if the determinant is negative then the roots will have opposite signs and your quadratic will be equivalent to xy = 1. But xy = 1 is not bounded in the xy plane, as you can see be substituting y = 1/x, for example, and getting x^2 + y^2 = 1/x^2 + x^2, which is not bounded as x goes to 0.Summary: Showing a set is bounded.In summary, to
cbarker1
Gold Member
MHB
TL;DR Summary
Showing a set is bounded.
Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1

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cbarker1 said:
Summary: Showing a set is bounded.

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \\ x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1
What about using ##(x -2y)^2 = x^2 -4xy + 4y^2##?

cbarker1
Now, how I do use an inequality to eliminate the ##(x-2y)^2##? At the end, I just want this fact ##x^2+y^2\leq 5##.

cbarker1 said:
Now, how I do use an inequality to eliminate the ##(x-2y)^2##? At the end, I just want this fact ##x^2+y^2\leq 5##.
Why not rewrite the original inequality using that equation?

cbarker1 said:
Now, how I do use an inequality to eliminate the ##(x-2y)^2##? At the end, I just want this fact ##x^2+y^2\leq 5##.
PS in case you missed the point that ##(x-2y)^2 \ge 0##

cbarker1
cbarker1 said:
Summary: Showing a set is bounded.

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1

Complete the square in one of the variables:

$$3x^2 - 4xy + 5y^2 = 3\left(x - \frac23y\right)^2 + \left(5 - \frac 43\right)y^2$$

PeroK said:
PS in case you missed the point that ##(x-2y)^2 \ge 0##
I have been working on this problem for a couple of weeks. So I did not see that helpful fact. Thanks.

PeroK
That squares are non-negative is generally one of the most widely used tricks in mathematics!

malawi_glenn and cbarker1
cbarker1 said:
Summary: Showing a set is bounded.

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1
##3x^2-4xy+5y^2 =2x^2+(x-2y)^2+y^2\leq x^2+(x-2y)^2+y^2##
Since ##(x-2y)^2\geq 0## then ##x^2+y^2\leq 5 \implies \sqrt{x^2+y^2}\leq \sqrt{5}##.

cbarker1 said:
##3x^2-4xy+5y^2 =2x^2+(x-2y)^2+y^2\leq x^2+(x-2y)^2+y^2##
Since ##(x-2y)^2\geq 0## then ##x^2+y^2\leq 5 \implies \sqrt{x^2+y^2}\leq \sqrt{5}##.
$$3x^2-4xy+5y^2 =2x^2+(x-2y)^2+y^2 \ge x^2+ y^2$$

cbarker1
Another option is to set $(x,y) = r(\cos\theta, \sin\theta)$ and show that the inequality can be written as $$r^2 \leq \frac{5}{A - B\cos(2(\theta-\theta_0))}$$ for some $\theta_0$ where $A > B > 0$. Then $$x^2 + y^2 = r^2 \leq \frac{5}{A - B}.$$

Delta2
another soution, similar to others: the given expression equals 2(x-y)^2 + x^2 + 3y^2. but

x^2+y^2 ≤ 2(x-y)^2 + x^2 + 3y^2 ≤ 5, since squares are non negative.

cbarker1
In general, as I recall, there are basically 3 types of quadratic expressions in (at most) 2 variables, x^2 = 1, x^2+y^2 = 1, and xy = 1, all others being essentially equivalent to one of these, only one of which is bounded in the xy plane. So you have to detect which type is yours. or you may consider -x^2 - y^2 and x^2-y^2 as somewhat different, but since x^2-y^2 = (x-y)(x+y) = uv, and -x^2-y^2 = -(x^2+y^2), you see these are closely related to the first 3.

Now essentially you want to measure how much your example is dominated by the terms x^2 and y^2, or by the term xy, so from ax^2 + 2bxy + cy^2, you form ac-b^2. Then your example is bounded, i.e. looks like x^2 + y^2 = 1, if and only if this expression is positive. In your case a= 5, c = 3, and 2b = 4, so ac-b^2 = 11 > 0, and it is bounded.

The reason is that one can "diagonalize" any example so that it looks like ax^2 + cy^2, where a and c can be zero or positive or negative. This is done by writing the quadratic form as a symmetric matrix expression, and diagonalizing that matrix. (This is equivalent to completing the square, i.e. getting rid of the coefficient of xy, as discussed above).

In your case the matrix has rows [5 -2] and [-2 3], and the determinant, ac-b^2 = 11. When your matrix is diagonalized, the entries on the diagonal, the new a and c, will be the roots of the characteristic polynomial of this (original) matrix, and their product will equal the determinant of this (original) matrix. So if that determinant is positive, then the roots will have the same sign and your quadratic will be equivalent to either x^2+y^2 or -(x^2+y^2), and both x^2+y^2 = 1 and -(x^2+y^2) = 1, are bounded, (since empty implies bounded).

To summarize, unless your quadratic form ax^2 + 2bxy + cy^2 actually defines a (doubled) straight line, or the empty set, it is equivalent to either x^2 + y^2, or x^2 - y^2, and you can tell which by computing ac-b^2. Indeed you only need the sign, which may be much easier than the number itself. E.g. if the coefficients a and c have opposite sign, you know ac-b^2 is negative, as in the case of say 3789x^2 - 5613 xy - 8201 y^2.

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dextercioby and cbarker1
A different approach is to do a translation/rotation of the right type, to make the xy-term disappear. of course, this will still require additional work, but it can give you insights. This case looks like you have a rotated ellipse. Rotations are isometries, meaning they preserve the distances between points, so set is bounded iff its rotated image is bounded. Basically, your new coordinates (x',y') will be rotated images of (x,y) , and then you set things so that the term x'y' is 0. This is a similar idea to that in mathwonk's post #13 above, right before this one.

mathwonk
Indeed the nice discussion in #14 gives a geometric way to view the algebraic "diagonalization" process referenced in # 13. I.e. to diagonalize, one finds the 2 eigenvalues and 2 corresponding eigenvectors of the symmetric matrix. It follows that the two eigenvectors are perpendicular, so the change of basis matrix they define (the matrix that diagonalizes the original one) can be chosen to be a rotation (take eigenvectors of length one and in the appropriate order).

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WWGD
nuuskur said:
Note that
$$3x^2 - 4xy + 5y^2 = 2(x-y)^2 + x^2 +3y^2 \geqslant 0.$$
Put ##f(x,y)=3x^2-4xy+5y^2##. One has ##S=f^{-1}([0,5])##. Preimage of closed and bounded (i.e compact) under continuous map is also closed and bounded. Polynomials are smooth (hence continuous).
The domain of ##f## is not compact.

nuuskur
nuuskur said:
Preimage of closed and bounded (i.e compact) under continuous map is also closed and bounded. Polynomials are smooth (hence continuous).
Take ##f(x)=5## for ##x\in\mathbb R##, it is continuous but the preimage of the closed and bounded set ##\{5\}## is the whole of ##\mathbb R## which is not bounded.

nuuskur

this is an easy mistake. the concept of a map for which the preimage of a compact set is also compact is very important in algebraic geometry and called a "proper map". if X and Y are compact Hausdorff spaces, then all continuous maps X-->Y are proper (and closed).

On metric spaces proper is equivalent to being a closed map with the inverse image of every point compact, neither of which is generally true for all continuous maps.

the definition of proper used in algebraic geometry is in terms of the forward properties of maps and emphasizes the fact that for say locally compact hausdorff spaces, proper is equivalent to being "universally closed", i.e. not only is X->Y closed, but so is XxZ-->YxZ for all Z.

Many of us have made the slip of failing to assume properness to derive some nice property of maps, including (me and) even the famous algebraic geometer I. Shafarevich, whose argument for the corollary of theorem 7 p. 61 (upper semi continuity, on the base, of fiber dimension) in his wonderful book Basic Algebraic Geometry, 1974 edition, is lacunary for this reason.

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nuuskur and martinbn

## What does it mean for a set to be bounded?

For a set to be bounded, it means that there exists a finite number M such that all elements in the set are less than or equal to M. This is known as an upper bound. Similarly, there exists a finite number m such that all elements in the set are greater than or equal to m, known as a lower bound.

## How do you show that a set is bounded?

To show that a set is bounded, you must find both an upper and lower bound for the set. This can be done by finding the maximum and minimum values in the set, or by using mathematical techniques such as the squeeze theorem or the triangle inequality.

## Can a set be bounded from above but not from below?

Yes, a set can be bounded from above but not from below. This means that there exists a finite number M such that all elements in the set are less than or equal to M, but there is no finite number m such that all elements in the set are greater than or equal to m.

## What is the difference between a bounded set and a finite set?

A bounded set refers to the upper and lower limits of the set, while a finite set refers to the number of elements in the set. A set can be bounded without being finite, and vice versa. For example, the set of all real numbers between 0 and 1 is bounded, but infinite.

## Can a set be bounded if it contains infinite elements?

Yes, a set can be bounded even if it contains infinite elements. For example, the set of all real numbers between 0 and 1 is bounded, but infinite. This is because there exists an upper bound of 1 and a lower bound of 0 for all elements in the set.

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