The difference between the limits of two Cauchy Sequences

[ANphysics]

Lets say that we have two Cauchy sequences {f_i} and {g_i} such that the sequence {f_i} converges to a limit F and the sequence {g_i} converges to a limit G. Then it can easily be shown that the sequence defined by
{ d(f_i, g_i) } is also Cauchy.

But is it true that this sequence, { d(f_i, g_i) }, converges to the limit d(F,G) ?

 

[HallsofIvy]

Yes, it is. If $\{f_i\}$ converges to F then, given any $\epsilon> 0$, there exist N such that if n> N then $|f_n- F|< \epsilon/2$. If $\{g_i\}$ converges to G then, given any $\epsilon> 0$, there exist M such that if n> M, then $|g_n- G|< \epsilon/2$.

Given any $\epsilon> 0$ if n> the larger of M and N, we have $|(f_i- g_i)- (F- G)|\le |f_i- F|+ |g_i- G|< \epsilon$.

 

[ANphysics]

But aren't you assuming here that distance is defined to be the absolute value of the difference between two elements. Is this true in general for any metric space?

For example, we can consider the vector space of all functions defined on [0,1] and define an inner product to be < f(x), g(x)> = ∫f(x)g(x)dx and a distance function to be
d(f(x),g(x))= (∫(f-g)(f-g)dx)^1/2. So in general when you have a metric space, but you don't necessarily know what that distance function is, can this statement be derived from the definition of metric space?

 

[Mandelbroth]

But aren't you assuming here that distance is defined to be the absolute value of the difference between two elements. Is this true in general for any metric space?

For example, we can consider the vector space of all functions defined on [0,1] and define an inner product to be < f(x), g(x)> = ∫f(x)g(x)dx and a distance function to be
d(f(x),g(x))= (∫(f-g)(f-g)dx)^1/2. So in general when you have a metric space, but you don't necessarily know what that distance function is, can this statement be derived from the definition of metric space?

Let's consider a discrete metric, which I shall denote r(x,y), and the case that $F=G$, but $f_i\neq g_i$. Then, the limit of $\{r(f_i,g_i)\}$ converges to 1, but $r(F,G)=0$.

As a fun little exercise, consider what happens when the metric is continuous. :tongue:

 

[jbunniii]

Let's consider a discrete metric, which I shall denote r(x,y), and the case that $F=G$, but $f_i\neq g_i$. Then, the limit of $\{r(f_i,g_i)\}$ converges to 1, but $r(F,G)=0$.

But if the metric is discrete, then $f_i \rightarrow F$ implies that $f_i = F$ for sufficiently large $i$, and similarly $g_i = G$ for sufficiently large $i$. Therefore we must eventually have $f_i = F = G = g_i$ if $i$ is large enough.

 

[jbunniii]

The question is whether the distance metric $d$ is a continuous function. This is true in any metric space. To see this, let $\epsilon > 0$, and note that $f_i \rightarrow F$ and $g_i \rightarrow G$ imply that for sufficiently large $i$, we have $d(f_i, F) < \epsilon/2$ and $d(g_i, G) < \epsilon / 2$.

Now we just play around with the triangle inequality to get what we need. First note the triangle inequality for real numbers gives us
$$|d(f_i, g_i) - d(F, G)| \leq |d(f_i, g_i) - d(f_i, G)| + |d(f_i, G) - d(F,G)|$$
But the triangle inequality for the metric $d$ gives us
$$|d(f_i, g_i) - d(f_i, G)| \leq d(g_i, G)$$
and
$$|d(f_i, G) - d(F,G)| \leq d(f_i, F)$$
so the first inequality becomes
$$|d(f_i, g_i) - d(F, G)| \leq d(f_i, F) + d(g_i, G)$$
which is bounded above by $\epsilon/2 + \epsilon/2 = \epsilon$ for sufficiently large $i$.

 

[jbunniii]

Metrics do not necessarily imply anything about a metric space. For example, consider using the discrete metric on the set of real numbers, since it can be applied to any nonempty set.

I guess I'm not seeing your point. The counterexample you offered does not work, for the reason I noted, regardless of what underlying set is given the discrete metric. Any convergent sequence under the discrete metric must be eventually constant. Therefore two convergent sequences with the same limit must be eventually equal to each other.

 

[WannabeNewton]

The only convergent sequences in a discrete topological (not just metric) space are the ones that are eventually constant. This is what my significant other jbun has pointed out.

The metric is certainly continuous on the product topology associated with the topology induced by the metric (the one generated by the base of open balls of the metric):
http://www.proofwiki.org/wiki/Metric_is_Continuous
http://math.stackexchange.com/quest...s-continuous-where-d-is-a-metric-defined-on-x

 

[Mandelbroth]

I guess I'm not seeing your point. The counterexample you offered does not work, for the reason I noted, regardless of what underlying set is given the discrete metric. Any convergent sequence under the discrete metric must be eventually constant. Therefore two convergent sequences with the same limit must be eventually equal to each other.

My apologies. I forgot that Cauchy sequences are defined by a metric when not working with real numbers. I was talking nonsense, so there was no point to see.

 

[WannabeNewton]

My significant other jbun says there is no need to apologize.

 

[jbunniii]

If I had a dollar for every time I've spoken nonsense, I could afford to buy WBN a pony.

 

[WannabeNewton]

I want a pony :[

 

[jbunniii]

You'll have to settle for a brony for now: http://www.brony.com/

 

[WannabeNewton]

Do the girls come with the shirts as a package deal? xP