Limit (sequence)

Note: see Limit page for other kinds of limits

Definition

Given a sequence $(x_n)_{n=1}^\infty\subseteq X$, a metric space $(X,d)$ (that is complete) and a point $x\in X$, the sequence $(x_n)$ is said to:

• have limit $x$ or converge to $x$

When:

• $$\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(x,x_n)<\epsilon]$$

  (note that $\epsilon\in\mathbb{R}$, obviously - as the co-domain of $d$ is $\mathbb{R}$)

• Read this as:

  for all $\epsilon$ greater than zero, there exists an $N$ in the natural numbers such that for all $n$ that are also natural we have that:

  whenever $n$ is beyond $N$ that $x_n$ is within $\epsilon$ of $x$

Equivalent definitions

Note: where it is not obvious changes have a $\{$ underneath them

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$$\lim_{n\rightarrow\infty}(x_n)=x\iff\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}\left[\underbrace{n\ge N}\implies d(x_n,x)<\epsilon\right]$$

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Here we have two definitions

1. $$\lim_{n\rightarrow\infty}(x_n)=x\iff\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}\left[n> N\implies d(x_n,x)<\epsilon\right]$$ (given at the top of the page)
2. $$\lim_{n\rightarrow\infty}(x_n)=x\iff\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}\left[n\ge N\implies d(x_n,x)<\epsilon\right]$$

Proof: $1\implies 2$

Let $\epsilon >0$ be given.

We know $\exists N'\in\mathbb{N}$ such that $n>N'\implies d(x_n,x)<\epsilon$ by assuming $1$ is true

Choose $N=N'+1$

Now $n\ge N\implies[n>N\vee n=N]$ by definition of $\ge$, substituting $N=N'+1$ we get $n\ge N'+1\implies[n>N'+1\vee n=N'+1]$

• (Case: $n>N'+1$) Note that $n>N'+1>N'$ so by transitivity of $>$ we see $n>N'$

  We know from $1$ that $n>N'\implies d(x_n,x)<\epsilon$

• (Case: $n=N'+1$), trivially $N'+1>N'$ so we have so $n>N'$

  We know from $1$ that $n>N'\implies d(x_n,x)<\epsilon$

So in either case we have $d(x_n,x)<\epsilon$

• We have shown that if $n\ge N$ we have $d(x_n,x)\epsilon$

• Thus choosing $N=N'+1$ is "an $N$ that exists" for the given $\epsilon$

This completes the first part of the proof

Proof: $2\implies 1$

Let $\epsilon>0$ be given.

We know $\exists N'\in\mathbb{N}$ such that $n\ge N'\implies d(x_n,x)<\epsilon$ by assuming $2$ is true

Choose $N=N'-1$

Now $n> N\implies n>N'-1$

• (Case: $n=N'$) if this is the case we know that $N'>N'-1$ so $n>N$ is satisfied, but also so is $n\ge N'$ (we have equality)

  We know from $2$ that this $\implies d(x_n,x)<\epsilon$

• (Case: $n>N'$) well $n\ge N'$ means "if $n>N'$ or $n=N'$" so $n>N'\implies n\ge N'$, thus $n\ge N'$

  We know from $2$ that this $\implies d(x_n,x)<\epsilon$

Thus for $n>N$ we see that $d(x_n,x)<\epsilon$

This completes the proof

(End of proof)

Discussion

Requiring $x\in X$

If $x\notin X$ then $d(x_n,x)$ is undefined, as $d:X\times X\rightarrow\mathbb{R}_{\ge_0}$, that is the distance metric is only defined for things in $X$.

To sidestep this limitation and talk about sequences that would converge if only their limit was in the space we consider Cauchy sequences. It is easy to see that all convergent sequences are Cauchy:

Cauchy sequence

Recall a Cauchy sequence is defined as:
Given a metric space $(X,d)$ and a sequence $(x_n)_{n=1}^\infty\subseteq X$ is said to be a Cauchy sequence if:

• $\forall\epsilon > 0\exists N\in\mathbb{N}\forall n,m\in\mathbb{N}[n\ge m> N\implies d(x_m,x_n)<\epsilon]$

In words it is simply:

• For any arbitrary distance apart, there exists a point such that any two points in the sequence after that point are within that arbitrary distance apart.

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Every convergent sequence is Cauchy

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Every convergent sequence is Cauchy/Proof

Process

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Discussion of why the definition is what it is.

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The idea is that defining "tends towards $x$" is rather difficult, to sidestep this we just say "we can get as close as we like to" instead. This is the purpose of $\epsilon$.

We say that "if you give me an $\epsilon>0$ - as small as you like - I can find you a point of the sequence ($N$) where all points after are within $\epsilon$ of $x$ (where $d(\cdot,\cdot)$ is our notion of distance)

• That is after $N$ in the sequence, so that's $x_{n+1},x_{n+1},\ldots$ the distance between $x_{N+i}$ and $x$ is $<\epsilon$

  This is exactly what $n>N\implies d(x_n,x)<\epsilon$ says, it says that:

  • whenever $n>N$ we must have $d(x_n,x)<\epsilon$

As per the nature of implies we may have $d(x_n,x)<\epsilon$ without $n>N$, it is only important that WHENEVER we are beyond $N$ in the sequence that $d(x_n,x)<\epsilon$

Example

Here: $x$-axis scale is from $0$ to $12.6$, marks are shown every unit. $y$-axis scale starts from $0$ and is marked every $0.25$ units. The sequence is any sequence of points on the wavy function shown. The limit of this is clearly $1$ The two horizontal lines show $1-\epsilon$ and $1+\epsilon$ The vertical line shows one possible value where every point after it is within $\epsilon$ of $1$ due to technical limitations the function $f(x)=1+\frac{\sin(\pi x)}{\frac{1}{4}x^2}$ is shown The curves are bounds on the function.

Notice that at $x=1$ that , in fact the curve is within $\pm\epsilon$ several times before we reach the vertical line, this is the significance of the implies sign, when we write $A\implies B$ we require that whenever $A$ is true, $B$ must be true, but $B$ may be true regardless of what $A$ is.

Note that after the vertical line the function is always within the bounds.

Because of this any $N'>N$ may be used too, as if $n>N'$ and $N'>N$ then $n>N'>N$ so $n>N$ - this proves that if $N$ works then any larger $N'$ will too. There is no requirement to find the smallest $N$ that'll work, just an $N$ such that $n>N\implies d(x_n,x)<\epsilon$

See also

• Cauchy sequence
• Limit