Exercise 13.6

Question

Show that the topologies of $\mathbb{R}_\ell$ and $\mathbb{R}_K$ are not comparable.

Amit Awasthi · Accepted Answer

\begin{proof}
  $\mathbb{R}_\ell 
ot\subseteq \mathbb{R}_K$. For $[a,b) \in \mathbb{R}_\ell$, there is no basis element $U \in \mathbb{R}_K$ such that $a \in U, U \subseteq [a,b)$. 
  \par $\mathbb{R}_K 
ot\subseteq \mathbb{R}_\ell$. For $(-1,1) \setminus K \in
  \mathbb{R}_K$ which contains $0$, there is no basis element $[a,b) \in
  \mathbb{R}_\ell$ such that $0 \in [a,b), [a,b) \subseteq (-1,1) \setminus K$
  by the Archimedean property, that is, for all $\epsilon > 0$, there exists $N
  \in \mathbb{N}$ such that $1/N < \epsilon$.
\end{proof}

Dan Whitman · Answer

Let $\mathcal{T}_l$ and $\mathcal{T}_K$ be the topologies of ${\mathbb{R}}_l$ and ${\mathbb{R}}_K$, respectively. Also let $\mathcal{B}_l$ and $\mathcal{B}_K$ be the corresponding bases. Consider $x = 0 \in {\mathbb{R}}$ and $B_l = [0,1)$, which clearly contains $0$ and is a basis element of $\mathcal{B}_l$. Let $B_K$ be any basis element of $\mathcal{B}_K$ that contains $0$. Then $B_K$ is either $(a,b)$ or $(a,b) - K$ for some $a < b$. In either case it must be that $a < 0 < b$ so that clearly $a < a/2 < 0 < b$. Also $a/2 otin K$ since $a/2 < 0$ so that we have $a/2 \in (a,b)$ and $a/2 \in (a,b) - K$. Clearly also $a/2 otin [0,1)$ so that it cannot be that $B_K \subset B_l$. We have therefore shown that \begin{gather*} \exists x \in {\mathbb{R}} \exists B_l \in \mathcal{B}_l \left[x \in B_l \land \forall B_K \in \mathcal{B}_K \left(x \in B_K \Rightarrow B_K ot \subset B_l ight) ight] \ \exists x \in {\mathbb{R}} \exists B_l \in \mathcal{B}_l \left[x \in B_l \land \forall B_K \in \mathcal{B}_K \left(x otin B_K \lor B_K ot \subset B_l ight) ight] \ \exists x \in {\mathbb{R}} \exists B_l \in \mathcal{B}_l \left[x \in B_l \land \lnot \exists B_K \in \mathcal{B}_K \left(x \in B_K \land B_K \subset B_l ight) ight] \ \exists x \in {\mathbb{R}} \exists B_l \in \mathcal{B}_l \lnot \left[x otin B_l \lor \exists B_K \in \mathcal{B}_K \left(x \in B_K \land B_K \subset B_l ight) ight] \ \exists x \in {\mathbb{R}} \exists B_l \in \mathcal{B}_l \lnot \left[x \in B_l \Rightarrow \exists B_K \in \mathcal{B}_K \left(x \in B_K \land B_K \subset B_l ight) ight] \ \lnot \forall x \in {\mathbb{R}} \forall B_l \in \mathcal{B}_l \left[x \in B_l \Rightarrow \exists B_K \in \mathcal{B}_K \left(x \in B_K \land B_K \subset B_l ight) ight] \ \lnot \forall x \in {\mathbb{R}} \forall B_l \in \mathcal{B}_l \left[x \in B_l \Rightarrow \exists B_K \in \mathcal{B}_K \left(x \in B_K \subset B_l ight) ight] \end{gather*} This shows by the negation of Lemma~13.3 that $\mathcal{T}_K$ is not finer than $\mathcal{T}_l$. Now consider again $x = 0 \in {\mathbb{R}}$ and $B_K = (-1,1) - K$, which clearly contains $0$ and is a basis element of $\mathcal{B}_K$. Let $B_l$ be any basis element of $\mathcal{B}_l$ that contains $0$ so that $B_l = [a,b)$ where $a \leq 0 < b$. Clearly we have that $1/b > 0$ and there is an $n \in {{\mathbb{Z}}_+}$ where $n > 1/b$ since the positive integers have no upper bound. We then have \begin{align*} 0 &< 1/b < n \ 0 &< 1 < bn & ext{(since $b > 0$)} \ 0 &< 1/n < b & ext{(since $n > 1/b > 0$)} \end{align*} so that $1/n \in [0,b) = B_l$. However, clearly $1/n \in K$ so that $1/n otin (-1,1) - K = B_K$. Hence it must be that $B_l ot \subset B_K$. This shows that $\mathcal{T}_l$ is not finer than $\mathcal{T}_K$ by the negation of Lemma~13.3 as before. This completes the proof that $\mathcal{T}_K$ and $\mathcal{T}_l$ are not comparable. \end{proof}

Exercise 13.6

Answers

Comments

Comments

Exercise 13.6

Answers

Comments

Comments

Add answer