# Complex numbers sequences/C is a metric space

## 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?

PeroK
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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?

Can you show that the two definitions of the metric are the same?

fishturtle1
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?

PeroK
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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?

PeroK
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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.

Thank you for the replies,
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

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## ...

PeroK
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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 ...?

PeroK
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... I had a geometric image of the number line with the number 2 displaced upwards. What would the metric for that set be?

... 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.

PeroK
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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.

fishturtle1
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..

PeroK
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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.

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 im 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##.

PeroK
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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.

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.