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I am asked to prove that

Show that a sequence $x: \mathbb{N} \to \mathbb{R}$ has a limit point iff there $\lim_{n\to\infty} x(n)$ exists as a limit point of a function from a subset of metric space $(\mathbb{N}*, d)$ to metric space $\mathbb{R}$. Note $\mathbb{N}* = \mathbb{N} \cup \infty$ and $d(m, n) = \vert \frac{1}{m} - \frac{1}{n} \vert$.

I am confused about what this problem is saying. How can a function have a limit point? Aren't those just for sets and sequences?

Thank you!

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The image of a function is a set. –  The Chaz 2.0 Oct 24 '11 at 17:51
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I believe that a better formulation would be limit of a function instead of limit point of a function. (With leaving the original intentions of your homework assignment.) –  Martin Sleziak Oct 24 '11 at 18:24
    
@MartinSleziak: Yeah, that's what my friend told me when I asked him. Thanks! –  badatmath Oct 24 '11 at 19:46
    
@okay-at-math Glad to see your username at math.stackexchange has changed :-) –  Martin Sleziak Oct 24 '11 at 19:54
    
@MartinSleziak: I gained a level :D –  badatmath Oct 24 '11 at 20:34

3 Answers 3

up vote 1 down vote accepted

I believe the following picture might help to get the intuition. (At least for people familiar with the topological spaces.)

Convergent sequence as a function

This picture shows the space $\mathbb N\cup\{\infty\}$ (the sets on the picture form the base of the topology).

Let us define the function $f: \mathbb N\cup\{\infty\} \to \mathbb R$ by $f(n)=x_n$ and $f(\infty)=x$. (The function from the above picture.)

The following conditions are equivalent:

  • $\lim_{n\to\infty} x_n=x$;

  • the function $f$ is continuous;

  • $\lim_{n\to\infty} f(n)=x$ holds.

(The same is true, if $\mathbb R$ is replaced by any topological space.)

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Holy jeezus you made me a picture. Thank you :D –  badatmath Oct 26 '11 at 4:24
    
I had the picture before (for another purposes), I just copied it here. (Well, converted from eps and copied.) –  Martin Sleziak Oct 26 '11 at 6:47

A function certainly can have a limit as the variable approaches a certain quantity. We have from very early on dealt with such things as $\displaystyle\lim_{x\to 2} \:\:x^3$ and, more interestingly, $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}.$

Showing, for example, that $\displaystyle\lim_{x\to 2} \:\:x^3=8$ involves, essentially, showing that as the distance of $x$ from $2$ approaches $0$, the distance of $x^3$ from $8$ approaches $0$.

We also have a familiar definition of what it means for the sequence $x(1), x(2), x(3), \dots$ to have a limit $b$ as $n \to \infty$. Let's try to recast this definition in terms of distance. Does $\displaystyle\lim_{n\to \infty} x(n)=b$ mean that as the distance of $n$ from "$\infty$" approaches $0$, the distance of $x(n)$ from $b$ approaches $0$?

That sounds kind of absurd! As $n$ increases without bound, how can its "distance" from $\infty$ be said to approach $0$? First of all, there is no such thing as $\infty$. Secondly, even if we "add" an abstract object $\infty$ to the set of integers, the distance of any $n$ from $\infty$ will always be infinite.

However, that's for the ordinary intuitive notion of distance. Add an "ideal" object $\infty$ to $\mathbb{N}$, to make a new set $\mathbb{N}^\ast$. On this set, define a metric as in your post, except that we need to add that the distance from $\infty$ to any ordinary positive integer $n$ is $\frac{1}{n}$. You can verify that this indeed gives a metric on $\mathbb{N}^\ast$. Note also informally that as $n$ gets large, its distance from $\infty$ under this metric approaches $0$.

You are being asked to show that the sequence $x(1), x(2), \dots$ has limit $b$ in the ordinary sense iff the limit of the function $x$, as the distance of $n$ from $\infty$ approaches $0$, is $b$. In the part after the iff "distance" is in the sense of the metric that has just been defined on $\mathbb{N}^\ast$.

Writing out the solution is essentially a careful translation job.

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Thank you, that was a very clear explanation. –  badatmath Oct 26 '11 at 4:26

Yes Functions can have limit points in Metric Space. It is the same as functions having a limit in an Euclidean Space. Here it is just abstracted and is applied over a Metric Space.

I think this Link will answer your question. Limit of a Sequence

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I believe Raman means this sentence from the wikipedia article: "The limit of a sequence of points $\left(x_n:n\in \mathbb{N}\right)$ in a topological space T is a special case of the limit of a function: the domain is $\mathbb{N}$ in the space $\mathbb{N} \cup \lbrace +\infty \rbrace$ with the induced topology of the affinely extended real number system, the range is T, and the function argument n tends to $+\infty$, which in this space is a limit point of $\mathbb{N}$. " –  Martin Sleziak Oct 24 '11 at 18:13
    
Yeah, my friend told me that when they said "limit point," it was kind of like "limit." Thank you for the clarification :) –  badatmath Oct 26 '11 at 4:27

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