ANA1_L10

What does ${\lim }_{x\to p}f(x)$ mean?

First stab at a definition

Definition 1(Definition).

Let $f:\mathbb{R}\to \mathbb{R}$.
${\lim }_{x\to p}f(x)=q$ means for any $ϵ>0$, there exists a $\delta >0$, such that $0<|x-p|<\delta$ implies $|f(x)-q|<ϵ$.

Note that if we edited the definition to say $\delta \ge 0$ such that $|x-p|\le \delta$, $f(p)$ would (always) be a limit, as we can have $\delta =0$. Not a good idea. Besides, we do not want to make any assertion about $f(p)$.

Also note that we do NOT want to say $0<|f(x)-q|<ϵ$, since $f(x)$ may be a constant function around $x=p$.

What's wrong with this?

We would like to take the limit of a function as $x$ approaches a point $p$ which is not in its domain. Thus, we would like to allow the domain of the function to be something that does not contain $p$, i.e, some subset $E\subset \mathbb{R}$. However, we cannot just replace the domain $\mathbb{R}$ with $E$ in the definition, as this might result in the “such that” clause being vacuously true for some $p$ (when $E$ is a singleton set $\{p\}$, for example, ${\lim }_{x\to p}=q$ is true for all $q\in \mathbb{R}$), making the definition useless. Thus, we need to be more discerning about the types of subsets $E$ we allow in the definition.

Second stab at a definition

Definition 2(Definition).

Let $f:E(\subset \mathbb{R})\to \mathbb{R}$, such that $E$ contains a deleted neighborhood of $p$.
Then, ${\lim }_{x\to p}f(x)=q$ means for any $ϵ>0$, there exists a $\delta >0$, such that $0<|x-p|<\delta$ implies $|f(x)-q|<ϵ$.

Note that we have only modified the prerequisites. We have replaced $\mathbb{R}$ with a subset of $E\subset \mathbb{R}$, and required $E$ to contain a deleted neighborhood of $p$. We say deleted neighborhood since we do not want the membership of $p$ in $E$ to play any role in the definition.

This definition attempts to ensure that the “such that” clause is never vacuously true. If $E$ contains a (deleted) neighborhood of $p$, $E$ contains an open ball (an open interval in $\mathbb{R}$) centered at $p$ of some finite radius (minus $p$, won’t say it again). Thus, for any $\delta >0$, we will have $x$ satisfying $0<|x-p|<\delta$, so the “such that” clause will never be vacuously true.

What's wrong (or rather inconvenient) about this?

Consider a function $f:\mathbb{Q}(\subset \mathbb{R})\to \mathbb{R}$ defined as $f(x)=0$. While $\mathbb{Q}$ does not contain a deleted neighborhood of, say, $p=1$, we would still like to be able to say ${\lim }_{x\to 1}f(x)=0$. What we would actually like to mandate is regardless of how small $\delta$ becomes, we will always have some $x$ satisfying $0<|x-p|<\delta$.

Brutus’s stab at a definition

We formalize the requirements we have distilled so far by defining limit points.

Limit points

Rudin, 2.18

Definition 3(Definition).

Given $E\subset \mathbb{R}$, a point $p\in \mathbb{R}$ is a limit point of $E$, if for all $\delta >0$, $(p-\delta ,p+\delta )\cap E$ contains a point other than $p$.
We denote the set of limit points of $E$ by $E^{\prime }$.

Definition 4(Definition).

If $p\in E$ and $p\notin E^{\prime }$, $p$ is called an isolated point of $E$.

Note that the membership of $p$ in $E^{\prime }$ has got nothing to do with the membership of $p$ in $E$.

• $p\in E$, $p\in E^{\prime }$: For example, $1\in (0,2)$
• $p\in E$, $p\notin E^{\prime }$: $p$ is an isolated point of $E$. For example, $1\in \{1,2\}$
• $p\notin E$, $p\in E^{\prime }$: For example, $0\notin (0,1)$
• $p\notin E$, $p\notin E^{\prime }$: For example, $2\notin (0,1)$

Observe that if $E$ has a limit point, it cannot be of finite cardinality.

For $f:E(\subset \mathbb{R})\to \mathbb{R}$, it makes sense to define ${\lim }_{x\to p}f(x)$ only if $p$ is a limit point of the domain of $f$.

Definition of limit of a function in R

Definition 5(Definition).

Let $f:E(\subset \mathbb{R})\to \mathbb{R}$, and let $p$ be a limit point of $E$.
Then, ${\lim }_{x\to p}f(x)=q$ means for any $ϵ>0$, there exists a $\delta >0$, such that $0<|x-p|<\delta$ implies $|f(x)-q|<ϵ$.

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Theorems about limit points

Rudin, 2.20

Theorem 6(Theorem).

If $p$ is a limit point of a set $E$, then every neighborhood of $p$ contains infinitely many points of $E$.

Fairly easy to see.

Rudin, 3.2 d

Theorem 7(Theorem).

If $X$ is a metric space, $E\subset X$ and if $p$ is a limit point of $E$, then there is a sequence $(p_n)$ in $E$ such that $p={\lim }_{n\to \infty }{p}_n$.

Ditto.

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Analogs for metric spaces

Definition of limit point in metric spaces

Definition 8(Definition).

Given $E\subset X$, a point $p\in X$ is a limit point of $E$, if for all $\delta >0$, $B_{\delta }(p)\cap E$ contains a point other than $p$.

Definition of limit of a function in metric spaces

Rudin, 4.1

Definition 9(Definition).

Let $X$ and $Y$ be metric spaces, $E\subset X$, $f:E\to Y$, $p$ is a limit point of $E$.
We write ${\lim }_{x\to p}f(x)=q$ , $q\in Y$ if for any $ϵ>0$ there exists a $\delta >0$ such that $0<d_X(x,p)<\delta$ implies $d_Y(f(x),q)<ϵ$, i.e, $f(B_{\delta }(p,E)\setminus \{p\})\subset B_{ϵ}(q,Y)$.

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Epilogue

We stated Rudin, 4.2, the sequence criterion.