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Topology problem sheet 1

 
  • Jan 16, 2023
Let $𝒯_\text{left}$ be the family of subsets $U$ of ℝ with the property that $βˆ€x∈UβˆƒΞ΅>0$ such that $(x-Ξ΅,x]βŠ‚U$.
Prove that $𝒯_\text{left}$ is a topology on ℝ.
What is the closure of $(0,1)$ with respect to this topology?
(T1) is trivial.
(T2) Let $U,Vβˆˆπ’―_\text{left}$. Consider an arbitrary $x∈U∩V$.
As $Uβˆˆπ’―_\text{left},βˆƒΞ±_x>0$ such that $(x-Ξ±_x,x]βŠ‚U$. As $Vβˆˆπ’―_\text{left},βˆƒΞ²_x>0$ such that $(x-Ξ²_x,x]βŠ‚V$.
Then $(x-Ξ΄_x,x]βŠ‚U∩V$ for $Ξ΄_x=\min(Ξ±_x,Ξ²_x)$. So $U∩Vβˆˆπ’―_\text{left}$.
(T3) Let $U_iβˆˆπ’―_\text{left}$. Consider an arbitrary $xβˆˆβ‹ƒ_{i∈I}U_i,βˆƒk∈I:x∈U_k$.
As $U_kβˆˆπ’―_\text{left}$, $βˆƒΞ΅_x>0:(x-Ξ΅_x,x]βŠ†U_kβŠ†β‹ƒ_{i∈I}U_i$. So $⋃_{i∈I}U_iβˆˆπ’―_\text{left}$.
Therefore $𝒯_\text{left}$ is a topology on ℝ.
If $1βˆˆβ„βˆ–\overline{(0,1)}$, then $βˆƒΞ΅>0:(1-Ξ΅,1]βŠ‚β„βˆ–\overline{(0,1)}$, but $1-\fracΞ΅2∈(1-Ξ΅,1]∩(0,1)$, contradiction.
So $1∈\overline{(0,1)}β‡’(0,1]βŠ‚\overline{(0,1)}$.
Also $β„βˆ–(0,1]βˆˆπ’―_\text{left}$, so $\overline{(0,1)}βŠ‚(0,1]$. Therefore $\overline{(0,1)}=(0,1]$.

Closure, interior, accumulation points

Let $(X, d)$ be a metric space and $A$ a non-empty subset of $X$.
For every $x∈X$, define the distance from $x$ to $A, \operatorname{dist}(x, A)$, by \[ \operatorname{dist}(x, A)=\inf_{a∈A} d(x, a) . \] Prove that $x∈\bar{A}$ iff $\operatorname{dist}(x, A)=0$.
See A2 sheet 2 Q5
$x∈\bar{A}β‡’βˆ€Ξ΅>0,βˆƒa∈A:d(x,a)<Ξ΅β‡’βˆ€Ξ΅>0:\operatorname{dist}(x, A)≀Ρ⇒\operatorname{dist}(x, A)=0$.
Conversely, $\operatorname{dist}(x, A)=0β‡’βˆ€Ξ΅>0:\operatorname{dist}(x, A)<Ξ΅β‡’βˆƒa∈A:d(x,a)<Ξ΅β‡’x∈\bar A$.
Let $(X, d)$ be a metric space.
  1. Prove that a closed ball is a closed set in $\left(X, 𝒯_d\right)$, where $𝒯_d$ is the topology defined by the metric $d$. Prove that the closure of an open ball is contained in the closed ball.
  2. Draw the picture of the closed ball with centre $\left(\frac12, 0\right)$ and of radius $\frac12$ in $\left(ℝ^2,{β€–β‹…β€–}_∞\right)$.
  3. Consider the subset $E=([0,1]Γ—\{0\})βˆͺ(\{0\}Γ—[0,1])$ of $ℝ^2$ endowed with the metric induced from $\left(ℝ^2,{β€–β‹…β€–}_∞\right)$. What are the open and the closed ball of centre $\left(\frac12, 0\right)$ and of radius $\frac12$ in this new metric space? Show that the closure of the open ball is a proper subset of the closed ball.
  4. Let $(V,{β€–β‹…β€–})$ be a normed real vector space. Show that in this case the closure of any open ball is a closed ball.
  1. A2 metric spaces Lemma 4.1.4
    Let the ball be $\bar{B}(a, Ξ΅)$. We will show that the complement $\bar{B}(a, Ξ΅)^c$ is open. Let $x∈\bar{B}(a, Ξ΅)^c$. Then $d(x, a)>Ξ΅$, so there is $Ξ΅'>0$ so that $d(x, a)>Ξ΅+Ξ΅'$. We claim that the open ball $B\left(x, Ξ΅'\right)$ is contained in $\bar{B}(a, Ξ΅)^c$. To see this, suppose that $z∈B\left(x, Ξ΅'\right)$. Then $d(z, x)< Ξ΅'$. By triangle inequality $d(z, a) β‰₯ d(x, a)-d(z, x)>\left(Ξ΅+Ξ΅'\right)-Ξ΅'=Ξ΅$.
    Since $B(a,Ξ΅)βŠ‚\bar{B}(a, Ξ΅)$ and we proved $\bar{B}(a, Ξ΅)$ is closed, by definition of closure, $\overline{B(a,Ξ΅)}βŠ‚\bar{B}(a, Ξ΅)$.
  3. Let $x=\left(\frac12,0\right)$, then $B\left(x,\frac12\right)=(0,1)Γ—\{0\},\overline{B\left(x,\frac12\right)}=[0,1]Γ—\{0\},\bar B\left(x,\frac12\right)=([0,1]Γ—\{0\})βˆͺ\left(\{0\}Γ—\left[0,\frac12\right]\right)$.
  4. When is the closure of an open ball equal to the closed ball?
    For any open ball $B(a, r)$ and $a+v∈\bar B(a,r)$, we have ${|v|}≀r$. Let $0<Ο΅<1$, then $a+Ο΅v∈B(a,r)$ and $a+Ο΅vβ†’a+v$ as $Ο΅β†’1$, so $a+v∈\overline{B(a,r)}$, so $\bar B(a,r)βŠ‚\overline{B(a,r)}$, but we proved $\bar B(a,r)βŠƒ\overline{B(a,r)}$, so $\bar B(a,r)=\overline{B(a,r)}$.
Let $f: X β†’ Y$ be a continuous map between topological spaces, and let $A$ and $B$ be subsets of $X$ such that $\bar{A}=\bar{B}$. Prove that $\overline{f(A)}=\overline{f(B)}$.
By Proposition 1.36, $\overline{f(A)}βŠƒf(\bar A)$.
Since $\overline{f(A)}$ is closed, we have $\overline{f(A)}βŠƒ\overline{f(\bar A)}$.
Also $\overline{f(A)}βŠ‚\overline{f(\bar A)}$, so $\overline{f(A)}=\overline{f(\bar A)}$.
Similarly $\overline{f(B)}=\overline{f(\bar B)}$, since $\bar{A}=\bar{B}$, we have $\overline{f(A)}=\overline{f(B)}$.
Give examples of subsets $A_1, A_2$ and $A_3$ of $ℝ^2$ (with its standard topology) such that
  1. $A_1'=ℝ^2$ and $\left(ℝ^2βˆ–A_1\right)'=ℝ^2$;
  2. $A_2'βŠ„A_2$ and $A_2βŠ„A_2'$;
  3. $A_3'''β‰ A_3''β‰ A_3'β‰ A_3$
  1. $A_1=β„š^2$. Since β„š is dense in ℝ, $A_1'=ℝ^2$. Similarly $\left(ℝ^2βˆ–A_1\right)'=ℝ^2$.
  2. $A_2=(0,1)^2βˆͺ\{(2,2)\}$, then $A_2'=[0,1]^2$, so $A_2'βŠ„A_2$ and $(2,2)βˆ‰A_2'$ so $A_2βŠ„A_2'$.
  3. Let $a_i(iβˆˆβ„•)$ be a sequence in ℝ. There exist balls $B_i$ centred at $a_i$ such that they are all disjoint.
    Let $a_{i,j}(jβˆˆβ„•)$ be sequence in $B_iβˆ–\{a_i\}$ converging to $a_i$. There exist balls $B_{i,j}$ centred at $a_{i,j}$ such that they are all disjoint.
    Let $a_{i,j,k}(kβˆˆβ„•)$ be a sequence in $B_{i,j}βˆ–\{a_{i,j}\}$ converging to $a_{i,j}$. There exist balls $B_{i,j,k}$ centred at $a_{i,j,k}$ such that they are all disjoint.
    Let $a_{i,j,k,l}(lβˆˆβ„•)$ be a sequence in $B_{i,j,k}βˆ–\{a_{i,j,k}\}$ converging to $a_{i,j,k}$.
    Then $A_3=\{a_{i,j,k,l}:i,j,k,lβˆˆβ„•\},A_3'=\{a_{i,j,k}:i,j,kβˆˆβ„•\},A_3''=\{a_{i,j}:i,jβˆˆβ„•\},A_3'''=\{a_i:iβˆˆβ„•\}$ are distinct.
    Iteration of derived sets
Give an example of an infinite topological space $X$ and a subset $AβŠ†X$, such that for any convergent sequence in $A$ any limit of it is in $A$, but $A$ is not closed.
Let $X=ℝ$ with co-countable topology, i.e. $A$ is open iff $A=βˆ…$ or $β„βˆ–A$ is at most countable. Then $A≔(-∞,0]$ contains all its limit points because for any sequence $x_n$ in $A$ and point $y>0$, the set $β„βˆ–\{x_n∣nβˆˆβ„•\}$ is an open neighbourhood of $y$ not containing any $x_n$, so any $yβˆ‰A$ is not a limit point of $A$. But $A$ is not closed because $β„βˆ–A=(0,+∞)$ is not countable.
Limit Points of Countable Complement Space
If a set contains all its limit points must it be closed?
Convergent sequence in co-finite topology iff there is at most one value occurs ∞ times.
Convergent sequence in co-countable topology iff sequence is eventually constant

Hausdorff spaces

Let $(X, 𝒯)$ be a Hausdorff topological space and let $A$ be a non-empty subset of $X$.
  1. Prove that an open set $Uβˆˆπ’―$ has non-empty intersection with $A$ iff $U$ has non-empty intersection with $\bar{A}$.
  2. Prove that $x∈A'$ iff $x∈\overline{Aβˆ–\{x\}}$.
  3. Prove that $A'$ is closed in $X$.
  4. Prove that $(\bar{A})'=A'$. Deduce that $\left(A'\right)'βŠ‚A'$.
  1. Suppose $A∩Uβ‰ βˆ…$, since $A∩UβŠ‚\bar A∩U$, we have $\bar A∩Uβ‰ βˆ…$.
    Suppose $\bar A∩Uβ‰ βˆ…$, let $x∈\bar A∩U$. Since open set $U$ contains $x∈\bar A$, by Proposition 1.30, $U∩Aβ‰ βˆ…$.
  2. $x∈\overline{Aβˆ–\{x\}}⇔$for any open $Uβˆ‹x,U∩(Aβˆ–\{x\})β‰ βˆ…β‡”x∈A'$.
  3. $βˆ€xβˆ‰A'$, by (2) the open set $Xβˆ–\overline{Aβˆ–\{x\}}$ contains $x$, so $A'$ is closed.
  4. $AβŠ‚\bar Aβ‡’A'βŠ‚(\bar{A})'$. In the other direction, $βˆ€x∈(\bar{A})'$, for any open set $Uβˆ‹x:\bar A∩Uβˆ–\{x\}β‰ βˆ…$.
    $X$ is T1$β‡’\{x\}$ is closed$β‡’Uβˆ–\{x\}$ is open. By (1), $A∩Uβˆ–\{x\}β‰ βˆ…β‡’x∈A'β‡’(\bar A)'βŠ‚A'$. Therefore $(\bar A)'=A'$.
    $A'βŠ‚\bar Aβ‡’\left(A'\right)'βŠ‚(\bar A)'=A'$
Let $(X,𝒯)$ be a Hausdorff topological space and let $f: Xβ†’X$ be continuous.
Show that the set$\DeclareMathOperator{\fix}{Fix}$$$\fix(f)=\{x:f(x)=x\}$$is closed.
$F:Xβ†’XΓ—X;x↦(x,f(x))$ is continuous.
Since $X$ is Hausdorff, $\Delta=\{(x,x):x∈X\}$ is a closed set in $XΓ—X$.
So $\fix(f)=F^{-1}(\Delta)$ is closed.
Alternate method:
To prove $Xβˆ–\fix(f)$ is open, for any $xβˆ‰\fix(f)$ we need to find open $Uβˆ‹x$ such that $U∩\fix(f)=βˆ…$.
By Hausdorff, we can find disjoint open sets $U_xβˆ‹x$ and $U_{f(x)}βˆ‹f(x)$. By continuity, $f^{-1}(U_{f(x)})$ is open, so $U_x∩f^{-1}(U_{f(x)})$ is open. Clearly $x∈U_x∩f^{-1}(U_{f(x)})$. It remains to prove $\fix(f)∩U_x∩f^{-1}(U_{f(x)})=βˆ…$.
Let $y∈\fix(f)∩U_x∩f^{-1}(U_{f(x)})$, then $y∈U_x$ and $y=f(y)∈U_{f(x)}$, contradicting $U_x,U_{f(x)}$ are disjoint.