Complex numbers sequences/C is a metric space

In summary, the proof shows that for any constant c in the set of complex numbers, if the limit of a sequence x_n equals L, then the limit of the sequence cx_n equals cL. This is proven by considering two cases - when the distance function d(c,0) is not equal to 0 and when it is equal to 0. In both cases, it is shown that the limit of cx_n equals cL, thus proving the statement.
  • #1
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Homework Statement


If ##\lim_{n \rightarrow \infty} x_n = L## then ##\lim_{n\rightarrow\infty}cx_n = cL## where ##x_n## is a sequence in ##\mathbb{C}## and ##L, c \epsilon \mathbb{C}##.

Homework Equations


##\lim_{n\rightarrow\infty} cx_n = cL## iff for all ##\varepsilon > 0##, there exists ##N(\varepsilon)## such that ##n > N(\varepsilon)## such that ##d(x_n, L) < \varepsilon## where ##d## is the distance function on ##\mathbb{C}##.

The Attempt at a Solution


For the rest of this post let ##x,y \epsilon \mathbb{C}##.

My notes say this:
Consider metric space ##\mathbb{C}## with ##d(x,y) = \sqrt{(x-y)\overline{(x-y)}}##. ##d(x,y)## is the standard euclidean distance for ##\mathbb{R}^2##.

Can we factor constants from ##d(cx, cy)##? Yes:
##d(cx, cy) = \sqrt{(cx - cy)\overline{(cx-cy)}}##
##= \sqrt{c\overline{c}(x-y)\overline{(x-y)}}##
##= \sqrt{c\overline{c}}\sqrt{(c-y)\overline{(x-y)}}##
##= \vert c \vert d(x,y)## where ##\vert c \vert## is modulus of ##c##.

Note: ##\vert c\vert = d(c,0)##.
----End of notes-----Now we prove if ##\lim_{n\rightarrow\infty}x_n = L##, then ##\lim_{n\rightarrow\infty} cx_n = cL##.

Proof: By definition of limit, for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##n >N(\varepsilon)## implies ##d(x_n, L) < \varepsilon##. Suppose ##d(x_n, L)## is the standard Euclidean distance metric for ##\mathbb{R}^2##. Then ##d(cx_n, cL) = \vert c \vert d(x_n, L) < \vert c \vert \varepsilon##. But ##\varepsilon## is arbitrary. So ##\lim_{n\rightarrow\infty}cx_n = cL##. []

But what if ##d(x,y)## isn't the standard Euclidean distance metric for ##\mathbb{R}^2##? Also this ##d(x,y)## doesn't look standard to me? Why wouldn't it be ##d(a+bi, m+ni) = \sqrt{(a-m)^2+(b-n)^2}##?

What if ##d(x,y)## isn't the standard Euclidean distance metric for ##\mathbb{R}^2##? How do you tell how many distance functions there are for ##\mathbb{C}## and for sets in general?
 
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  • #2
fishturtle1 said:

Homework Statement


If ##\lim_{n \rightarrow \infty} x_n = L## then ##\lim_{n\rightarrow\infty}cx_n = cL## where ##x_n## is a sequence in ##\mathbb{C}## and ##L, c \epsilon \mathbb{C}##.

Homework Equations


##\lim_{n\rightarrow\infty} cx_n = cL## iff for all ##\varepsilon > 0##, there exists ##N(\varepsilon)## such that ##n > N(\varepsilon)## such that ##d(x_n, L) < \varepsilon## where ##d## is the distance function on ##\mathbb{C}##.

The Attempt at a Solution


For the rest of this post let ##x,y \epsilon \mathbb{C}##.

My notes say this:
Consider metric space ##\mathbb{C}## with ##d(x,y) = \sqrt{(x-y)\overline{(x-y)}}##. ##d(x,y)## is the standard euclidean distance for ##\mathbb{R}^2##.

Can we factor constants from ##d(cx, cy)##? Yes:
##d(cx, cy) = \sqrt{(cx - cy)\overline{(cx-cy)}}##
##= \sqrt{c\overline{c}(x-y)\overline{(x-y)}}##
##= \sqrt{c\overline{c}}\sqrt{(c-y)\overline{(x-y)}}##
##= \vert c \vert d(x,y)## where ##\vert c \vert## is modulus of ##c##.

Note: ##\vert c\vert = d(c,0)##.
----End of notes-----Now we prove if ##\lim_{n\rightarrow\infty}x_n = L##, then ##\lim_{n\rightarrow\infty} cx_n = cL##.

Proof: By definition of limit, for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##n >N(\varepsilon)## implies ##d(x_n, L) < \varepsilon##. Suppose ##d(x_n, L)## is the standard Euclidean distance metric for ##\mathbb{R}^2##. Then ##d(cx_n, cL) = \vert c \vert d(x_n, L) < \vert c \vert \varepsilon##. But ##\varepsilon## is arbitrary. So ##\lim_{n\rightarrow\infty}cx_n = cL##. []

But what if ##d(x,y)## isn't the standard Euclidean distance metric for ##\mathbb{R}^2##? Also this ##d(x,y)## doesn't look standard to me? Why wouldn't it be ##d(a+bi, m+ni) = \sqrt{(a-m)^2+(b-n)^2}##?

What if ##d(x,y)## isn't the standard Euclidean distance metric for ##\mathbb{R}^2##? How do you tell how many distance functions there are for ##\mathbb{C}## and for sets in general?

Can you show that the two definitions of the metric are the same?
 
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  • #3
PeroK said:
Can you show that the two definitions of the metric are the same?
I didn't realize that, yes:

Let ##x = a + bi## and ##y = m + ni## for some real numbers ##a,b,m,n##.

Then ##\sqrt{(x-y)\overline{(x-y)}} = \sqrt{(a + bi - m - ni)(a - bi + m +ni)}##
##= \sqrt{a^2 + b^2 - 2am - 2bm + m^2 + n^2}##
##= \sqrt{(a-m)^2 + (b-n)^2}##

Ok so now I'm clear on the distance function from class, but how come we don't need to consider any other distance function?
 
  • #4
fishturtle1 said:
I didn't realize that, yes:

Let ##x = a + bi## and ##y = m + ni## for some real numbers ##a,b,m,n##.

Then ##\sqrt{(x-y)\overline{(x-y)}} = \sqrt{(a + bi - m - ni)(a - bi + m +ni)}##
##= \sqrt{a^2 + b^2 - 2am - 2bm + m^2 + n^2}##
##= \sqrt{(a-m)^2 + (b-n)^2}##

Ok so now I'm clear on the distance function from class, but how come we don't need to consider any other distance function?

Analysis depends on the specific metric. Complex analysis usually means with the standard Euclidean metric.

Can you find an example of a metric where the original proposition does not hold? Or, does it hold for any metric?
 
  • #5
Ps I'd say your proof in the original post is more the outline of a proof. You should try to fill out the details.
 
  • #6
Thank you for the replies,
PeroK said:
Ps I'd say your proof in the original post is more the outline of a proof. You should try to fill out the details.
Let me try again

We'd like to show: If ##\lim_{n\rightarrow\infty} x_n = L## then ##\lim_{n\rightarrow\infty} cx_n = cL## where ##x_n## is a sequence in ##\mathbb{C}## and ##c, L \epsilon \mathbb{C}##.

Proof: Suppose ##\lim_{n\rightarrow\infty}x_n = L## and let ##c \epsilon \mathbb{C}##. By assumption, our distance function defined as ##d(x,y) = \sqrt{(x-y)\overline{(x-y)}}##.

Consider 2 cases.

case 1: ##d(c,0) \neq 0##. We have for all ##\varepsilon > 0## there exists ##N := N(\frac{\varepsilon}{d(c,0)})## such that ##n > N## implies ##d(x_n, L) < \frac{\varepsilon}{d(c,0)}##. Since ##d## is the standard Euclidean metric, we have ##d(cx_n, cL) = d(c,0)d(x_n,L) < \varepsilon## i.e., ##d(cx_n, cL) < \varepsilon##. Therefore ##\lim_{n\rightarrow\infty}cx_n = cL##.

case 2: ##d(c,0) = 0##. By property of distance function, ##d(c,0) = 0## iff ##c = 0##. Since ##\lim_{n\rightarrow\infty} 0 = 0##, we have ##\lim_{n\rightarrow\infty} cx_n = cL##.

Thus, for any constant ##c\epsilon\mathbb{C}##, if ##\lim_{n\rightarrow\infty} x_n = L## then ##\lim_{n\rightarrow\infty} cx_n = cL##. []
Are there things I could do to make this proof better/cleaner?

Still working on #4
 
  • #7
PeroK said:
Analysis depends on the specific metric. Complex analysis usually means with the standard Euclidean metric.

Can you find an example of a metric where the original proposition does not hold? Or, does it hold for any metric?
Just to make sure, the question is whether or not ##d(c,0)d(x,y) = d(cx, cy)## for any distance function that forms a metric space with ##\mathbb{C}##.

This holds at least for ##c = 0## because if c = 0 then ##d(c,0)d(x,y) = 0\cdot d(x,y) = 0## and ##d(cx, cy) = d(0\cdot x, 0\cdot y) = d(0,0) = 0## so ##d(c,0)d(x,y) = d(cx, cy)##.

Now suppose ##c \neq 0## ...
 
  • #8
fishturtle1 said:
Just to make sure, the question is whether or not ##d(c,0)d(x,y) = d(cx, cy)## for any distance function that forms a metric space with ##\mathbb{C}##.

This holds at least for ##c = 0## because if c = 0 then ##d(c,0)d(x,y) = 0\cdot d(x,y) = 0## and ##d(cx, cy) = d(0\cdot x, 0\cdot y) = d(0,0) = 0## so ##d(c,0)d(x,y) = d(cx, cy)##.

Now suppose ##c \neq 0## ...

The way I looked at it was. We might as well let ##c=2##, ##L = 1## and keep things real.

So, we have a sequence converting to 1, but we don't want double the sequence to converge to 2.

Can we construct a metric where 2 is special, or an anomaly in some way ...?
 
  • #9
... I had a geometric image of the number line with the number 2 displaced upwards. What would the metric for that set be?
 
  • #10
PeroK said:
... I had a geometric image of the number line with the number 2 displaced upwards. What would the metric for that set be?
What does 2 displaced upwards mean?

edit: I've ruled out any scalar multiple of absolute value, any kth root of ##x_n - L##, and ##d(x, y) \neq f(x,y) + k## where ##k## is some constant and ##f(x,y)## is some expression dependent on ##x## and ##y##, because they would never get within small epsilon##(<k)## of each other.
 
  • #11
fishturtle1 said:
What does 2 displaced upwards mean?

The number line with 2 at the position that the point (2,1) would be, say.

Remember that a metric is a "distance" function. I visualised 2 being a certain distance from the rest of the number line. I.e. it's further from all the other numbers than it normally is.

And by "normally" I mean with the usual metric.
 
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  • #12
PeroK said:
The number line with 2 at the position that the point (2,1) would be, say.

Remember that a metric is a "distance" function. I visualised 2 being a certain distance from the rest of the number line. I.e. it's further from all the other numbers than it normally is.

And by "normally" I mean with the usual metric.
Ok, so for any ##a \epsilon \mathbb{R}: d(a, 2) = \vert 2 - a \vert - 1 + \sqrt{2}##. So could a metric function be defined as for any ##a, b \epsilon \mathbb{R}: ( a \le 2 \le b \rightarrow d(a,b) = d(a,2) + d(2, b)) \wedge (a,b > 2 \vee a, b < 2 \rightarrow d(a,b) = \vert a - b \vert) ##?

I see that if ##\lim_{n\rightarrow\infty} x_n = 1##, then ##\vert 2x_n - 2\vert > -1 + \sqrt{2}## so ##\lim_{n\rightarrow\infty}2x_n \neq 2##.

edit: I sincerely appreciate your help on this, and I'm going to sleep on it and try again in the morning with a clear head..
 
  • #13
fishturtle1 said:
Ok, so for any ##a \epsilon \mathbb{R}: d(a, 2) = \vert 2 - a \vert - 1 + \sqrt{2}##. So could a metric function be defined as for any ##a, b \epsilon \mathbb{R}: ( a \le 2 \le b \rightarrow d(a,b) = d(a,2) + d(2, b)) \wedge (a,b > 2 \vee a, b < 2 \rightarrow d(a,b) = \vert a - b \vert) ##?

I see that if ##\lim_{n\rightarrow\infty} x_n = 1##, then ##\vert 2x_n - 2\vert > -1 + \sqrt{2}## so ##\lim_{n\rightarrow\infty}2x_n \neq 2##.

I don't follow all of that. More simply, you could define:

##d(2,x) = |2-x| +1## (for ##x \ne 2##)

Check that is a metric. Should be easy enough.

Then take any sequence converting to 1, such as ##1 - \frac1n##.

Remember that for a counterexample you need a specific sequence.

This geometric thinking should help you create your own counterexample in future.

Also, this shows that a metric is not necessarily determined by algebraic properties.
 
  • #14
PeroK said:
I don't follow all of that. More simply, you could define:

##d(2,x) = |2-x| +1## (for ##x \ne 2##)

Check that is a metric. Should be easy enough.

Then take any sequence converting to 1, such as ##1 - \frac1n##.

Remember that for a counterexample you need a specific sequence.

This geometric thinking should help you create your own counterexample in future.

Also, this shows that a metric is not necessarily determined by algebraic properties.
For the picture we have the real number line but at 2 there is a hole, and 2 is moved 1 unit upward, right?

So with this distance function, if ##x,y < 2## or ##x,y > 2##, then is ##d(x,y) = \vert x - y \vert##? Is the reasoning behind this metric to measure the distance from a point ##x## to ##2## and not worry about ##x## and some other point ##y##?

To check its a distance function:
1)##d(2,x) = \vert 2 - x \vert + 1 \ge 0## for all ##x \epsilon \mathbb{R}-\lbrace 2\rbrace##
2)##d(2,2) = 0## by the picture?
3)##d(x,2) = \vert x-2 \vert + 1 = \vert 2 - x \vert + 1 = d(2,x)## so ##d(x,2) = d(2,x)## for all ##x \epsilon \mathbb{R} - \lbrace 2\rbrace##.
4) Triangle inequality: ##d(x,2) + d(y,2) = \vert 2 - x \vert + 1 + \vert 2 - y \vert + 1## and ##d(x+y, 2) = \vert 2 - x - y \vert + 1##...?

Triangle inequality in general: ##d(x,z) \le d(x,y) + d(y,z)## for any ##x,y,z \epsilon X## where ##X## is some set. But here it seems we fix 2? So what is the triangle inequality with this metric?

Also for the sequence ##1 - \frac 1n##, ##d(1 - \frac 1n, 2) = \vert 2 - 1 + \frac 1n \vert + 1 = \vert 1 + \frac 1n \vert + 1##. So ##d(1 - \frac 1n, 2) \rightarrow 2## as ##n \rightarrow + \infty##.

But ##d(2\cdot (1 - \frac 1n), 2) \rightarrow 2## as ##n \rightarrow + \infty##. So ##\lim_{n\rightarrow\infty} 2\cdot(1 - \frac 1n) = 2## but I am not sure how to evaluate ##\lim_{n\rightarrow\infty} 1 - \frac 1n## with the current metric, all I can tell is it doesn't converge to 2. I know with the standard Euclidean metric, ##\lim_{n\rightarrow\infty} 1 - \frac 1n = 1##.
 
  • #15
You may not have understood. The metric is the same at the usual Euclidean metric on ##\mathbb{R}##, except where the number ##2## is concerned.

The only thing that really needs to be checked is The triangle inequality. But that comes out easily:

##d(2,x) = |2-x|+1 \le |2-y| +|y-x|+1 = d(2,y) +d(y,x)## etc.

We also take ##d(2,2) = 0##, by definition.

But, in this metric no number is closer than 1 unit to 2. So, the only sequences that converge to 2 must be constant after some point.

In particular, the sequence ##2 - \frac2n## has no limit with this metric.

But the sequence ##1 - \frac1n## converges to 1, as it's the same as in the usual metric.

In simple terms, all we've done is moved 2 away from the rest of the number line.

Perhaps don't spend too much time worrying about this. But, this is the sort of approach that is useful to generate a counterexample in many cases.
 
  • #16
PeroK said:
You may not have understood. The metric is the same at the usual Euclidean metric on ##\mathbb{R}##, except where the number ##2## is concerned.

The only thing that really needs to be checked is The triangle inequality. But that comes out easily:

##d(2,x) = |2-x|+1 \le |2-y| +|y-x|+1 = d(2,y) +d(y,x)## etc.

We also take ##d(2,2) = 0##, by definition.

But, in this metric no number is closer than 1 unit to 2. So, the only sequences that converge to 2 must be constant after some point.

In particular, the sequence ##2 - \frac2n## has no limit with this metric.

But the sequence ##1 - \frac1n## converges to 1, as it's the same as in the usual metric.

In simple terms, all we've done is moved 2 away from the rest of the number line.

Perhaps don't spend too much time worrying about this. But, this is the sort of approach that is useful to generate a counterexample in many cases.
Thank you for your time and patience with me on this thread. I think I've gotten what I wanted out of it, so ill mark this as solved.
 

1. What are complex numbers sequences?

Complex numbers sequences are a mathematical concept that involves a series of complex numbers arranged in a specific order. A complex number is a number that has both a real and imaginary component, expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit.

2. How are complex numbers sequences different from real number sequences?

Complex numbers sequences differ from real number sequences in that they involve both real and imaginary numbers, while real number sequences only involve real numbers. Additionally, complex numbers sequences follow different rules for addition, subtraction, multiplication, and division compared to real number sequences.

3. What is a metric space?

A metric space is a mathematical concept that defines a set of objects, such as complex numbers, and a distance function between those objects. This distance function measures the distance between any two objects in the set and follows a set of rules, such as the triangle inequality, to ensure the distance is always positive and symmetrical.

4. How is C a metric space?

C, or the set of complex numbers, is a metric space because it satisfies the properties of a metric space. It has a defined distance function, the modulus or absolute value, which follows the rules of a metric space. For any two complex numbers, the distance between them is always positive and symmetrical, and it satisfies the triangle inequality.

5. What are some applications of complex numbers sequences and metric spaces?

Complex numbers sequences and metric spaces have various applications in mathematics, physics, engineering, and computer science. They are used in signal processing, quantum mechanics, and electrical circuit analysis, as well as in computer graphics and data compression. They are also essential in the study of fractals and chaos theory.

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