ANA1_L13

Recall

Recall

• Definition of continuity at a point from L11
  • How it contrasts with the definition of limit
  • How metric space setup simplifies the language (Rudin, 4.12)
  • Continuity of limit points and isolated points
  • Continuity definition in terms of open balls
• Definition of continuity of a function from L11

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A digression

Had a rather lengthy discussion on why the definition of continuity as stated in L11 would remain unaltered if we edited $d_X(x,p)<\delta$ to say $0<d_X(x,p)<\delta$ instead.

The learning from the discussion was one pertaining to basic logic. Consider three logical statements $A$, $B$, and $C$. Define $P$ and $Q$ thus:

$$\begin{array}{r}P:A⟹B\\ Q:C⟹B\end{array}$$

Now, if $C⟹A$, what can we say about $P$ and $Q$? if $P$ is true, then $C⟹A⟹B$, i.e $Q$ is true. Thus, hypothesis of $Q$ $⟹$ hypothesis $P$ implies $P⟹Q$ and vice versa.

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What does it mean for a function to be continuous?

In L11, we defined $f$ to be continuous if $f$ was continuous at every point in its domain. Now, we want to define continuity of a function in terms of its codomain, to motivate the notion of open sets. But first, an observation.

A ball is a neighborhood of each of its points

Rudin, 2.19

Theorem 1(Theorem).

In a metric space $X$, any ball is a neighborhood of each point in it, i.e,

$$q\in B_r(p)⟹\exists s\text{ such that }{B}_s(q)\subset B_r(p)$$

Proof
Choose $s=r-d(p,q)$. Use triangle inequality. ❏

As defined later, this basically means every open ball is an open set.

If $f$ is continuous, $f^{-1}$ of any ball is a neighborhood of each of its points

Now, we make the following claim:

Theorem 2(Claim).

Given $f:X\to Y$, $f$ is continuous $⟹$ $f^{-1}(B)$ is a neighborhood of any of the points of $f^{-1}(B)$, where $B$ is a ball in $Y$.

Proof
$f^{-1}(B)$ contains all the point $x\in X$ such that $f(x)\in B$. From the previous theorem, there exists an $ϵ>0$ such that $B_{ϵ}(f(x))\subset B$. Since $f$ is continuous, there exists a $\delta >0$ such that $f(B_{\delta }(x))\subset B_{ϵ}(f(x))$, which implies $f(B_{\delta }(x))\subset B$, which implies $B_{\delta }(x)\subset f^{-1}(B)$. ❏

This property of “being a neighborhood of any of its points” is going to be the defining property of open sets. There exists another logically equivalent property: being a union of open balls.

Equivalence of the two properties

Theorem 3(Claim).

The following are equivalent for for any subset $S$ of a metric space $X$:

1. $S$ is a neighborhood of any $p\in S$ for all $p\in S$, i.e, for any $p\in S$ there exists $r>0$ such that $B_r(p)\subset S$.
2. $S$ is a union of open balls.

Proof of $1⟹2$
For every point $p\in S$, consider an open ball $B_{r_p}(p)$. Consider the union of all these balls. Since every $p$ has a ball representing it, $S\subset \bigcup B_{r_p}(p)$. Also, since every $B_{r_p}(p)\subset S$, it must be that $\bigcup B_{r_p}(p)\subset S$. Thus, $S=\bigcup B_{r_p}(p)$. ❏

Proof of $2⟹1$
We have already shown that every ball satisfies 1. It follows that a union of balls must also satisfy 1, since every point in the union will be a part of at least one ball. ❏

It follows from the first theorem that ($f$ is continuous $⟹$ $f^{-1}$ of any ball in $Y$ is a union of balls in $X$). Since $f^{-1}$ preserves unions (Abbot exercise 1.2.9, if you’re unsure), we can make the more general statement ($f$ is continuous $⟹$ $f^{-1}$ of union of balls in $Y$ is a union of balls in $X$). Having motivated it enough, we finally define an open set.

Open sets

Definition 4(Definition).

Subset $U$ of a metric space $X$ is called open $\boxed{\text{ in }X}$ if $U$ is a union of balls in $X$(or, equivalently, $U$ contains a ball in $X$ around each point in $U$).

Example 5((counter)Examples of open sets).

• Any open interval in $\mathbb{R}$ is an open set.
• $\mathbb{R}$ is open in $\mathbb{R}$.
• $\mathbb{R}$ is not open in $\mathbb{C}$.

Important

1. A set $U$ is open or not open only in the context of a superset $X$, and when this context changes, the status of $U$ is liable to change too. Hence the emphasis of $X$ in the definition. Check examples above.
2. Being open is NOT the negation of being closed or vice versa. The two properties exist independently of each other.

Halfway through to a new definition of continuity

The definition of open sets distills our previously stated implication of $f$ being continuous to
($f$ is continuous $⟹$ $f^{-1}$ of an open set in $Y$ is an open set in $X$).

Now, we put forth a stronger claim: $f$ is continuous $⟺$ $f^{-1}$(open set in $Y$) is an open set in $X$. Put more rigorously, this is Rudin, 4.8.

Rudin, 4.8

Theorem 6(Theorem).

A mapping $f:X\to Y$, where $X$ and $Y$ are metric spaces, is continuous on $X$ if and only if $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$.

Having already shown $⟹$, it only remains to show its converse.

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Epilogue

Defined closed sets.

Definition 7(Definition).

A subset $C\subset X$ is called closed $\boxed{\text{ in }X}$ if $X\setminus C$ is open in $X$ (or, equivalently, $C$ contains all the limit points of $C$).

Stated properties of open and closed sets under set operations:

• Any union of open sets is open (it it understood that we are working in the context of a fixed metric space). This should be easy to see.
  • If $U_1$ and $U_2$ are open, so is $U_1\bigcap U_2$. Hence any finite intersection of open sets is open. Prove this. << supposed to be easy
• Any intersection of closed sets is closed. (let $C_1$, $C_2$ be closed sets. $\overline{C_1}$ and $\overline{C_2}$ are open sets, so $\overline{C_1}\bigcup \overline{C_2}$ is open. Its complement, $C_1\bigcup C_2$, is closed.)
  • If $C_1$ and $C_2$ are closed, so is $C_1\bigcup C_2$. Hence any finite union of closed sets is closed. Prove this. << ditto