05. Open and Closed Sets

5.1 Basic Definitions

Open Sets

If $X$ is a Metric Space then subset $U\subseteq X$ is open if

For each $y\in U$ there exists some $\delta >0$ such that

$$\text{open ball }B(y,\delta )\text{ is contained in }U$$

Link to original

Open Balls in a Matric Space lemma

Every Open ball in a metric space is an Open Set

Proof

Let the ball be $B(a,ϵ)$
Let $x\in B(a,ϵ)$ then

$$d(x,a)<ϵ$$

So there exists $e^{\prime }>0$ such that

$$d(x,a)<ϵ-{ϵ}^{\prime }$$

Need to show that open ball $B(x,{ϵ}^{\prime })$ is contained in $B(a,ϵ)$

Suppose that $z\in B(x,{ϵ}^{\prime })$ then

$$d(z,x)<{ϵ}^{\prime }$$

So by Distance function - Triangle Inequality theqn

$$d(z,a)\le d(z,x)+d(x,a)<{ϵ}^{\prime }+(ϵ-{ϵ}^{\prime })=ϵ$$

So then we get that $B(x,{ϵ}^{\prime })\subseteq B(\alpha ,ϵ)$
Hence $B(a,ϵ)$ is an open set

Link to original

Closed Set

If $X$ is a Metric space then subset $F\subseteq X$ is closed if and only if

$$\text{complement }{F}^c=X\setminus F\text{ is an open subset}$$

Link to original

Complement of an Open Set

Complement of an Open Set is a Closed Set

Link to original

Property of Open and Closed Sets

In a Metric space, it is possible for a subset to be

1. open
2. closed
3. both
4. neither

Link to original

Closed Balls in a Metric Space lemma

Every closed ball in a metric space is a closed set
In particular, singleton sets are closed

Proof

Let ball be $\overline{B}(a,ϵ)$
Let $x\in \overline{B}(a,e{)}^c$ then

$$d(x,a)>ϵ$$

So there exists ${ϵ}^{\prime }>0$ such that

$$d(x,a)>ϵ+{ϵ}^{\prime }$$

Suppose $z\in B(x,{ϵ}^{\prime })$ then

$$d(z,x)<{ϵ}^{\prime }$$

So by Distance function - Triangle Inequality we have

$$d(z,a)\ge d(x,a)-d(z,x)>(ϵ+{ϵ}^{\prime })-{ϵ}^{\prime }=ϵ$$

Hence $B(x,{ϵ}^{\prime })\subseteq \overline{B}(a,ϵ{)}^c$
So $\overline{B}(a,ϵ)$ is a closed ball
Singleton sets are closed as $\{a\}=\overline{B}(a,0)$

Link to original

---

5.2 Basic Properties of Open Sets

Properties of Open Sets lemma

Let $X$ be a Metric space then

1. Subsets $X$ and $\varnothing$ are open
2. For any indexing set $I$ and $\{U_i:i\in I\}$ is a collection of open sets then

$$\bigcup _{i\in I}{U}_i\text{ is an open set}$$

3. If $I$ is finite and $\{U_i:i\in I\}$ are open sets then

$$\bigcap _{i\in I}{U}_i\text{ is open }$$

Proof

1. Trivial…

2. If $x\in {\bigcup }_{i\in I}{U}_i$ then there is some $i\in I$ with $x\in U_i$
   Since $U_i$ is open then there exists open ball $B(x,ϵ)$ contained in $U_i$ hence also

$$B(x,ϵ)\subseteq \bigcup _{i\in I}{U}_i$$

So ${\bigcup }_{i\in I}{U}_i$ is an open set
3) Suppose $I$ is finite and $x\in {\bigcap }_{i\in I}{U}_i$ then

$$x\in U_i$$

So as $U_i$ is open then there exists a open ball $B(x,{ϵ}_i)$ contained in $U_i$ for some ${ϵ}_i>0$
Let $ϵ:={\min }_{i\in I}{ϵ}_I$ then
As $I$ is finite then $ϵ>0$ so then

$$B(x,ϵ)\subseteq B(x,{ϵ}_i)\subseteq U_i\text{ for all }i\in I$$

Hence

$$B(x,ϵ)\subseteq \bigcap _{i\in I}{U}_i$$

Hence ${\bigcap }_{i\in I}{U}_i$ is an open set

Note that $(1)$ is just a special case of $(2)$ and $(3)$ where $I=\varnothing$

Link to original

Properties of Closed Sets lemma

Let $X$ be a Metric space then

1. Subsets $X$ and $\varnothing$ are closed
2. For any indexing set $I$ and $\{F_i:i\in I\}$ is a collection of closed sets then

$$\bigcup _{i\in I}{U}_i\text{ is an closed set}$$

3. If $I$ is finite and $\{U_i:i\in I\}$ are closed sets then

$$\bigcap _{i\in I}{U}_i\text{ is closed }$$

Proof - Small

Use proof for Properties of open sets by taking complements and de Morgan’s Laws

Link to original

Topology

Let $X$ be a metric space then

Topology of $X$ is the collection of all open sets in $X$

Link to original

---

5.3 Continuity in Terms of Open Sets

Continuity in terms of Open Sets

Let $X,Y$ be metric spaces and let $f:X\to Y$ be a map then

$$f\text{ is continuous on all of }X⟺\text{ for open subset }U\subseteq Y,\text{ preimage }{f}^{-1}(U)\text{ is open in }X$$

Proof

Proof $⟹$
Suppose $f$ is continuous at every point

Let $U\subseteq Y$ be open
Let $a\in f^{-1}(U)$ be arbitrary then $f(a)\in U$ so since $U$ is open then exists $ϵ>0$ such that

$$B(f(a),ϵ)\subseteq U$$

By definition of continuity, there exists some $\delta >0$ such that

$$x\in B(a,\delta )⟹f(x)\in B(f(a),ϵ)$$

Therefore

$$f^{-1}(B(f(a),ϵ))\supseteq B(a,\delta )$$

Hence $B(a,\delta )\subseteq f^{-1}(U)$ so $f^{-1}(U)$ is open

Proof $⟸$
Suppose $f$ satisfies the open set preimages property then let $a\in X$

Ball $B(f(a),ϵ)$ is open so by assumption then preimage $f^{-1}(B(f(a),ϵ))$ is open
As $a$ is in this set then by definition of open then there exists some $\delta >0$ such that

$$B(a,\delta )\subseteq f^{-1}(B(f(a),ϵ))$$

hence $f(B(a,\delta ))\subseteq B(f(a),ϵ)$ then it means $f$ is continuous at $a$

Link to original

Continuity in terms of Closed Sets

Let $X,Y$ be metric spaces and let $f:X\to Y$ be a map then

$$f\text{ is continuous on all of }X⟺\text{ for closed subset }V\subseteq Y,\text{ preimage }{f}^{-1}(V)\text{ is closed in }X$$

Proof

Similar to Continuity in terms of Open Sets Proof but take Complements

Link to original

---

5.4 Topological Spaces (Non-Examinable)

Ermmmm maybe another time 😭

Although refer to page $39$ of lecture notes

---

5.5 Subspaces

Balls of Subspaces

Let $(X,d)$ be a metric space then by $Y\subseteq X$ is a subspace then

$$B_Y(y,r)=\{z\in Y:d(z,y)<r\}$$

is for the open ball about $y$ of radius $r$ in $Y$

With notable property

$$B_X(y,r)=\{x\in X:d(x,y)<r\}$$

is for the open ball for radius $r$ about $y$ in $X$ with property then

$$B_Y(y,r)=Y\cap B_X(y,r)$$

Link to original

Property of Closed and Open Subsets between Subspaces lemma

Let $X$ be a metric space and suppose $Y\subseteq X$ then

$$U\subseteq Y\text{ is an open subset of }Y⟺\text{ exists open subset }V\subseteq X\text{ such that }U=Y\cap V$$

Similarly for closed subsets se we have

$$Z\subseteq Y\text{ is a closed subset of }Y⟺\text{ exists closed subset }F\subseteq X\text{ such that }Z=F\cap V$$

Proof

Proof $⟸$
Suppose that $U=Y\cap V$ where $V$ is open in $X$
Let $y\in U$ then since $V$ is open there exists $ϵ>0$ such that $B_X(y,ϵ)\subseteq V$ hence

$$B_Y(y,ϵ)=Y\cap B_X(y,ϵ)\subseteq Y\cap V=U$$

Hence $U$ is open

Proof $⟹$
Suppose that $U$ is an open subset of $Y$
For each $y\in U$ there exists ${ϵ}_y>0$ such that $B_Y(y,{ϵ}_y)\subseteq U$ then

$$\bigcup _{y\in U}{B}_Y(e,{ϵ}_y)=U$$

Define

$$V:=U_{y\in U}{B}_X(y,{ϵ}_y)$$

As $V$ is a union of open balls in $X$ then it is open with property

$$Y\cap V=Y\cap \bigcup _{y\in Y}{B}_X(y,{ϵ}_y)=\bigcup _{y\in Y}{B}_Y(y,{ϵ}_y)=U$$

For the closed sets results just take complements

Link to original