- Let $X=C[0,1]$, with the usual distance coming from the ${\|·\|}_∞$-norm. Is $X$ connected?
Proof.
For any $f,g∈X$, there is a continuous map $h:[0,1]→C[0,1],h(λ)=(1-λ)f+λg$ with $h(0)=f,h(1)=g$. So $X$ is path-connected. By Theorem 7.4.1. $X$ is connected. - a) Let $A⊆ℝ^2$ be the set of all points with at least one rational coordinate. $A$ is connected.
b) Let $A⊆ℝ^2$ be the set of all points with at least one irrational coordinate. $A$ is connected.
c) Let $A⊆ℝ^2$ be the set of all points with exactly one rational coordinate. $A$ is disconnected.
d) Let $A⊆ℝ^2$ be the set of all points with two rational coordinates. $A$ is disconnected.
e) Let $A⊆ℝ^2$ be the set of all points with two irrational coordinates. $A$ is disconnected.
Proof.
a) For any $(x,y)∈A$, wlog let $x$ be rational. Consider the polyline Γ from $(0,0)$ to $(x,0)$ to $(x,y)$. Any point on Γ has a rational coordinate, so $Γ⊂A$, so $A$ is a star domain, so $A$ is path connected, so $A$ is connected.
b) For any $(x,y)∈A$, wlog let $x$ be irrational. Consider the polyline Γ from $(0,π)$ to $(x,π)$ to $(x,y)$. Any point on Γ has an irrational coordinate, so $Γ⊂A$, so $A$ is connected.
c) Let $U=\{(x,y)∈ℝ^2∣x>y\}$ and $V=\{(x,y)∈ℝ^2∣x< y\}$. No element of $A$ lies on the line $x=y$, so $A⊂U∪V$. Since $U$ and $V$ are open sets, and $U∩V∩A=∅$, by lemma 7.1.3 $A$ is disconnected.
d) Let $U=\{(x,y)∈ℝ^2∣x>π\}$ and $V=\{(x,y)∈ℝ^2∣x< π\}$. No element of $A$ lies on the line $x=π$, so $A⊂U∪V$. Since $U$ and $V$ are open sets, and $U∩V∩A=∅$, by lemma 7.1.3 $A$ is disconnected.
e) Let $U=\{(x,y)∈ℝ^2∣x>0\}$ and $V=\{(x,y)∈ℝ^2∣x< 0\}$. No element of $A$ lies on the line $x=0$, so $A⊂U∪V$. Since $U$ and $V$ are open sets, and $U∩V∩A=∅$, by lemma 7.1.3 $A$ is disconnected. - Show that there is no continuous injective map $f:ℝ^2→ℝ$. [Hint: consider the restriction of $f$ to $ℝ^2∖\{a\}$, for a suitable point $a$]
Proof.
By Lemma 7.1.6 the image of a connected set $ℝ^2$ under continuous map $f$ is connected, that is a non-degenerate interval $I⊂ℝ$. Take an interior point $a$ of $I$, then $I∖\{a\}$ is not connected.
But $ℝ^2∖\{f^{-1}(a)\}$ is connected, by Lemma 7.1.6 the image under $f$ is connected, then $I∖\{a\}$ is connected, contradiction. - Let $X$ be a metric space and $A_1, A_2, …$ an infinite collection of subsets of $X$. For each of the following statements, give a proof or counterexample.
- If $A_1, A_2,…, A_k$ are sequentially compact then so is $A_1∪A_2∪…∪A_k$.
- If $A_1, A_2,…, A_k$ are connected then $A_1∩A_2∩…∩A_k$ is connected.
- If $A_1, A_2,…$ are sequentially compact then $⋃_{k⩾1} A_k$ is sequentially compact.
- If $A_1, A_2,…$ are connected and $A_j∩A_{j+1}≠∅$ then $⋃_{k⩾1} A_k$ is connected. [Generalization of Lemma 7.1.4]
-
Proof: Let $(x_n)$ be a sequence in $A_1∪A_2∪…∪A_k$.
$∃i∈\{1,2,⋯,k\}$, $A_i$ contains infinite many terms of $(x_n)$, so they form a subsequence $(y_n)$.
By sequential compactness of $A_i$, $(y_n)$ has a convergent subsequence, and it is a subsequence of $(x_n)$. Therefore $A_1∪A_2∪…∪A_k$ is sequentially compact. - Counterexample: $A_1=\{z∈ℂ:{|z|}=1,\operatorname{Im}z≥0\},A_2=\{z∈ℂ:{|z|}=1,\operatorname{Im}z≤0\}$, then $A_1∩A_2=\{1,-1\}$ is not connected.
- Counterexample: $A_k=[k-1,k]$ is sequentially compact. Then $⋃_{k⩾1}A_k=[0,+∞)$ is unbounded, so is not sequentially compact, by Proposition 8.5.1.
-
Proof: Suppose $⋃_{k⩾1}A_k=X_1∪X_2$ where $X_1,X_2$ are disjoint open sets in $X$. Since $A_k,k=1,2,⋯$ is connected, we have either $A_k⊂X_1$ or $A_k⊂X_2$.
If $A_1⊂X_1$, since $A_1∩A_2≠∅$, we have $A_2⊂X_1$, by the same argument $A_k⊂X_1,k=1,2,⋯$, so $X_2=∅$. Similarly if $A_1⊂X_2$, then $X_1=∅$. In conclusion $⋃_{k⩾1}A_k$ is connected.
- Show that ℤ with the 2-adic metric is not connected.
Proof.
\begin{array}l B(0,1)=\{y:d_2(0,y)< 1\}=\{y:2|y\}\\ B(1,1)=\{y:d_2(1,y)< 1\}=\{y:2|y-1\} \end{array}$B(0,1),B(1,1)$ are disjoint and open sets and $ℤ=B(0,1)∪B(1,1)$. So $ℤ$ is not connected with the 2-adic metric. - Suppose that $X$ is connected and that $f: X→Y$ is locally constant, meaning that every $x∈X$ lies in some open set $U$ on which $f$ is constant. Show that $f$ is constant.
Proof.
Take $x∈X$. Let $U=f^{-1}(f(x))$. Every $y∈U$ lies in some open set $U_y$ such that $∀z∈U_y:f(z)=f(y)⇒f(z)=f(x)⇒z∈U⇒U_y⊂U$. So $U$ is open.
Every $w∈X∖U$ lies in some open set $U_w$ such that $∀z∈U_w:f(z)=f(w)⇒f(z)≠f(x)⇒z∈X∖U⇒U_w⊂X∖U$. So $X∖U$ is open.
$U$ is open and closed, and nonempty(since $x∈U$), therefore $U=X$, in other words, $f$ is constant. - Suppose that $f:X→ℕ$ is continuous. Show that $f$ is constant on every connected component of $X$.
Proof 1.
Let $X_1$ be a connected component of $X$, then $X_1$ is connected. $∀x∈X_1$, Since $f$ is continuous, $∃δ>0,∀y∈B(x,δ):{|f(x)-f(y)|}< 1$, but $f(x),f(y)∈ℕ$, so $f(x)=f(y)$, so $f$ is locally constant on $X_1$. Applying Q6 we're done.
Proof 2.
Let $X_1$ be a connected component of $X$. Take an element $x_1$ of $X_1$. Let $U=\{x∈X_1:f(x)=f(x_1)\}$ and $V=\{x∈X_1:f(x)≠f(x_1)\}$.
Since $X_1$ is connected and $X_1=U∪V$ and $U≠∅$(as $x_1∈X_1$), we have $U=X_1$, so $f$ is constant on $X_1$. - Is there a metric on ℕ which makes it connected?
Solution.
No. Let $d$ be the supposed metric such that ℕ is connnected. For $x_0∈ℕ$, the function $Φ:ℕ→ℝ^+$ defined by $Φ(x)=d(x,x_0)$ satisfy ${|Φ(x)-Φ(y)|}≤d(x,y)$ by reverse triangle inequality, so Φ is continuous, so $Φ(ℕ)$ is connected. By positivity, $Φ(ℕ)$ has at least 2 elements (0 and a positive number), so $Φ(ℕ)$ contains a non-degenerate interval, so $Φ(ℕ)$ is uncountable, contradiction. - Let $(V,{\|·\|})$ be a normed vector space whose unit sphere $S=\{v ∈V:{\|v\|}=1\}$ is sequentially compact.
- Show that any closed ball $B=\{v∈V:{\|v\|}≤R\}$ is sequentially compact.
- Show that $V$ is complete.
- Proof 1. $∀\{x_n\}⊂B(0;R)$.
$r_n={‖x_n‖}$ is a sequence in $[0,R]$, so it has a subsequence $r'_n→r$. If $r=0$, then $x_n→0$; if $r≠0$, take $δ∈(0,r)$, then $∃N,∀n>N:r'_n>δ$.
Consider the map $(r,θ)↦rθ$ from $[δ,R]×S$ to $\{v∈V:δ≤{\|v\|}≤R\}$. Since continuous surjective maps preserve compactness, $(x_n')$ has a convergent subsequence.
Proof 2. $∀\{x_n\}⊂B(0;R)$. $r_n={‖x_n‖}$ is a sequence in $[0,R]$, so it has a subsequence $r'_n→r$. If $r=0$, then $x_n→0$; if $r≠0$, $∃N,∀n>N:r'_n≠0$.
Let $θ'_n=\frac{x'_n}{r'_n}$, then $θ'_n∈S$, so it has a subsequence $θ''_n→θ$.
So $x''_n=r''_nθ''_n→rθ$. So $(x''_n)$ is a convergent subsequence of $(x_n)$. - Let $(x_n)$ be a Cauchy sequence in $V$, then $(x_n)$ is bounded, so $(x_n)$ is contained in a closed ball $B(0,R)$ for some $R$.
By a) $B(0,R)$ is compact, by Proposition 8.5.1. $B$ is complete, so $(x_n)$ converges, so $V$ is complete.
- Let $X$ be a subset of $ℝ^n$ such that every continuous function $f: X → ℝ$ is bounded. Show that $X$ is sequentially compact.
Proof.
If $X$ is unbounded, then $f:x↦{\|x\|}$ is unbounded and continuous, which contradicts our assumption. So $X$ is bounded.
If $X$ is closed, by Bolzano-Weierstrass, $X$ is sequentially compact.
If $X$ is not closed, $∃x∈\overline X∖X$, then $ϕ:t↦{\|x-t\|}$ is continuous on $X$ and $ϕ(t)≠0$. By Sheet 1 Q6, $1/ϕ$ is continuous on $X$.
$x∈\overline X⇒∀M>0,∃t∈X∩B(x,1/M)⇒1/ϕ(t)>M$. Therefore $1/ϕ$ is unbounded on $X$. - Let $\|⋅\|$ be an arbitrary norm on $ℝ^n$. Show that there is some constant $C$ such that
${\|v\|}≤C{\|v\|}_1$ for all $v ∈ ℝ^n$ Using this, show that there is some constant $c>0$ such that ${\|v\|}⩾ c{\|v\|}_1$ for all $v ∈ℝ^n$.
Proof. [lecture notes Proposition 3.3.2]
Step 1: We wish to show that $$ c{\|x\|}_1≤{\|x\|}≤C{\|x\|}_1, $$ is true for all $x ∈V$ for some $c, C$. It is trivially true for $x=0$, so we need only consider $x≠0$, in which case we can divide by ${\|x\|}_1$ to obtain the condition $$ c≤{\|u\|}≤C, $$ where $u=x /{\|x\|}_1$ satisfies ${\|u\|}_1=1$. So only need consider $\left\{u:{\|u\|}_1=1\right\}$.
Step 2: We wish to show ${\|⋅\|}: ℝ^n → ℝ$ is continuous function on $ℝ^n$ with the ${\|⋅\|}_1$-norm. That is, $∀ϵ>0, ∃δ>0, \left\|x-x'\right\|_1< δ ⇒{\big|{\|x\|}-\left\|x'\right\|\big|}< ϵ$.
By the triangle inequality $$ {\big|{\|x\|}-\left\|x'\right\|\big|}≤\left\|x-x'\right\| $$ Applying the triangle inequality again, and writing $x=\sum_{i=1}^{n} \alpha_i e_i$ and $x'=\sum_{i=1}^{n} \alpha_i' e_i$ in our basis, we obtain $$ \left\|x-x'\right\|≤\sum_{i=1}^{n}\left|\alpha_i-\alpha_i'\right| ·\left\|e_i\right\|≤\max _i\left\|e_i\right\|⋅\sum_{i=1}^{n}\left|\alpha_i-\alpha_i'\right|=\max _i\left\|e_i\right\|⋅\left\|x-x'\right\|_1 $$ Therefore, if we choose $$ δ=\frac{ϵ}{\max _i\left\|e_i\right\|} $$ it immediately follows that $$ \left\|x-x'\right\|_1< δ ⟹{\big|{\|x\|}-\left\|x'\right\|\big|}≤\left\|x-x'\right\|< ϵ. $$ Step 3: By the extreme value theorem(or Lemma 8.3.1: a continuous image of a sequentially compact set is sequentially compact), a continuous function $\|·\|$ on compact set $\left\{u:{\|u\|}_1=1\right\}$ can achieve a maximum and minimum in the set. Let \begin{aligned} c&=\min_{\|u\|_1=1}{\|u\|} \\ C&=\max_{\|u\|_1=1}{\|u\|} \end{aligned} Since $u≠0$ for ${\|u\|}_1=1$, it follows that $C≥c>0$ and $$ c≤{\|u\|}≤C $$ as required by step 1. Q.E.D. - Write down an infinite compact subset of ℚ and prove that it is compact directly from the open cover definition of compactness.
Solution.
$X=\{0\}\cup\left\{\frac1n:n∈ℤ^+\right\}$ is an infinite subset of ℚ. Let $𝒰=\{U_i:i∈I\}$ be an open cover of $X$.
Let $\frac1n∈X$ be covered by $U_{i_n}∈𝒰$. Let $0∈X$ be covered by $U_{i_0}∈𝒰$. Since $U_{i_0}$ is open, $∃ϵ>0,B(0,ϵ)⊂U_{i_0}$.
Let $r=\left⌊\frac1ϵ\right⌋$, then $X∖B(0,ϵ)=\left\{\frac1n:n=1,⋯,r\right\}$ is finite (empty if $r=0$). Therefore $\{U_{i_n}:n=0,1,⋯,r\}$ is a finite subcover. - Consider the space Ω of all sequences $𝐱=(x_n)_{n=1}^∞$ with $x_n∈[0,1]$ for all $n$, together with the metric $d(𝐱,𝐲)=\sum_{k=1}^∞ 2^{-k}\left|x_k-y_k\right|$. Show that Ω is sequentially compact.
Proof 1.
For a sequence in Ω $$𝐗_0 = 𝐱_1, 𝐱_2,…$$ we need to extract a convergent subsequence. Let $$𝐗_1 = 𝐱_{11}, 𝐱_{12}, …$$ be a subsequence of $𝐗_0$ such that the first coordinate of each sequence $𝐱_{11}, 𝐱_{12}, …$ converges. This is possible from sequential compactness of $[0,1]$.
Similarly, for all $n≥2$, let $𝐗_n$ be a subsequence of $𝐗_{n-1}$ such that the $n$-th coordinate converges.
Now consider the sequence $$𝐘 = 𝐱_{11}, 𝐱_{22}, 𝐱_{33}, … $$ formed by taking the diagonal of the array \begin{array}l 𝐱_{11}& 𝐱_{12}& 𝐱_{13}&…\\ 𝐱_{21}& 𝐱_{22}& 𝐱_{23}&…\\ 𝐱_{31}& 𝐱_{32}& 𝐱_{33}&…\\ ⋮&⋮&⋮&⋱ \end{array} We claim this sequence converges in Ω. Note that for $m< n$, $𝐱_{nn}$ is an element of $𝐗_m$, and it comes strictly after $𝐱_{mm}$ in $𝐗_m$. Thus, $$ 𝐱_{11}, 𝐱_{22}, … $$ is a subsequence of $𝐗_1$, so 𝐲 converges in the first coordinate. Similarly, $$ 𝐱_{22}, 𝐱_{33}, … $$ is a subsequence of $𝐗_2$, so 𝐲 converges in the second coordinate. Continuing like this, we see that 𝐲 converges pointwise in all coordinates, the rest is same as Proof 2.(b)(ii) Proof 2.
To prove Ω is compact, by Theorem 8.5.3. we need to prove (a) Ω is totally bounded (b) Ω is complete.
(a) $∀ϵ>0,∃m∈ℕ:\sum_{k≥m}2^{-k}<\fracϵ2$. Let $F_ϵ$ be a finite subset of $[0,1]$ such that the open $\fracϵ2$-balls centred at elements of $F_ϵ$ cover $[0,1]$. For each $k≥m$ let $x_k∈[0,1]$ be arbitrary. Let $F$ be the set of all sequences $(a_k)$ such that $a_k∈F_ϵ$ for $k< m$ and $a_k=0$ for $k≥m$. Clearly $F$ is a finite subset of $X$. Now we prove $ϵ$-balls centred at elements of $F$ cover Ω.
For any $𝐲=(y_k)_{k∈ℕ}∈Ω$. Construct $𝐱=(x_k)_{k∈ ℕ}∈F$ by $∀k< m,∃x_k∈F_ϵ:y_k∈B\left(x_k,\fracϵ2\right)$ and $∀k≥m:y_k=0$. Then \begin{align*}d(𝐱,𝐲)&=\sum_{k∈\Bbb N}2^{-k}\left|x_k-y_k\right|\\ &=\sum_{k< m}2^{-k}\left|x_k-y_k\right|+\sum_{k≥m}2^{-k}\left|x_k-y_k\right|\\ &<\sum_{k< m}\fracϵ{2^{k+1}}+\sum_{k≥m}2^{-k}\\ &<\fracϵ2+\fracϵ2\\ &=ϵ \end{align*} Therefore $X$ is totally bounded.
(b) For every Cauchy sequence $(𝐱_k)_{k ∈ℕ} = \left( \left( x_k^{(j)} \right)_{j ∈ℕ} \right)_{k ∈ℕ}$, we do as follows: (i) $∀j_0 ∈ℕ$, the $j_0$ th term $\left( x_k^{(j_0)} \right)_{k ∈ℕ}$ converges to $y^{(j_0)}$, (ii) set $𝐲=\left( y^{(j)} \right)_{j ∈ℕ}$, prove $\lim_{k→∞}𝐱_k = 𝐲$.
(i) fix $j_0 ∈ℕ, ϵ > 0$. By the Cauchy assumption, $∃K = K(2^{-j_0}ϵ) ∈ℕ$ such that if $k_1, k_2≥K$, then $$d(𝐱_{k_1} ,𝐱_{k_2} ) = \sum_{j = 1}^∞ 2^{-j} \left| x_{k_1}^{(j)} - x_{k_2}^{(j)} \right| < 2^{-j_0}ϵ.$$LHS includes the term $2^{ - j_0} \left| x_{k_1}^{(j_0)} - x_{k_2}^{(j_0)} \right|$, so $\left| x_{k_1}^{(j_0)} - x_{k_2}^{(j_0)} \right|< ϵ$, so $\left( x_k^{(j_0)} \right)_{k ∈ℕ}$ is Cauchy, and thus has a limit, call it $y^{(j_0)}$.
(ii) Now we show that $𝐲 = \lim_{k→∞}𝐱_k$. For every $j ∈ℕ , η > 0$, $∃L(j, η)$ such that if $k≥L(j, η)$, then $\left| x_k^{(j)} - y^{(j)} \right| < η$.
Fix $ϵ> 0$. $∃J ∈ℕ$ such that $\sum_{j = J + 1}^∞ 2^{-j} < ϵ/ 2$. Let $L = \max \left\{ L \left( 1, ϵ/ 2 \right) , … , L \left( J, ϵ/ 2 \right) \right\}$. Suppose $k≥L$. Then \begin{align*} d(𝐱_k , 𝐲 ) & = \sum_{j = 1}^∞ 2^{-j} \left| x_k^{(j)} - y^{(j)} \right| \\ & = \sum_{j = 1}^{J} 2^{-j} \left| x_k^{(j)} - y^{(j)} \right| + \sum_{j = J + 1}^∞ 2^{-j} \left| x_k^{(j)} - y^{(j)} \right| \\ & < \sum_{j = 1}^{J} 2^{-j} (ϵ/ 2) + \sum_{j = J + 1}^∞ 2^{-j} \\ & < (ϵ/ 2) \sum_{j = 1}^∞ 2^{-j} + ϵ/ 2 \\ & = ϵ/ 2 + ϵ/ 2 \\ & = ϵ. \end{align*}