- #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?