Return to Answer

What I realized from Mr. Ozawa's comment is as follows:

Step1. There exists an $\epsilon>0$ and a sequence $\{x_n\}\subset T$ such that the open balls $\{U_\epsilon(x_n)\}_{n\in\mathbb{N}}$ are disjoint.

This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup_{n=0}^NU_\epsilon(x_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup_{n=0}^NU_\epsilon(x_n)$, we have $$d(x,\{x_n\}_{n=0}^N)<2\epsilon.$$ This implies that the family $$\{U_{2\epsilon}(x_n)\}_{n=0}^N$$ covers $T$, contradicting the assumption that $T$ is totally bounded.

Step2. There is a Banach space embedding $l^\infty\hookrightarrow C_u(T)$.

By defining $f_n(x):=\min\left\{1-\frac{d(x,x_n)}{\epsilon},0\right\}$ for all $n\in\mathbb{N}$, there exist functions $\{f_n\}\subset C_u(T;[0,1])$ such that $$f_n(x_n)=1,\quad f_n|_{T\setminus U_\epsilon(x_n)}=0,\quad n\in\mathbb{N}.$$ Using these, we can define the following bounded linear operator: $$(a_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}a_nf_n$$ This operator preserves the norm, so it is also a Banach space embedding.

Step3. $l^\infty$ is not separable.

The set of indicator functions $\{\chi_I\}_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.

Separable spaces do not have a non-separable subspace, so this constitutes a contradiction.

If there are no serious mistakes, I would like to post the same answer on the original query on Mathematics Stach Exchange.

Thank you all so much.

What I realized from Mr. Ozawa's comment is as follows:

Step1. There exists an $\epsilon>0$ and a sequence $\{x_n\}\subset T$ such that the open balls $\{U_\epsilon(x_n)\}_{n\in\mathbb{N}}$ are disjoint.

This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup_{n=0}^NU_\epsilon(x_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup_{n=0}^NU_\epsilon(x_n)$, we have $$d(x,\{x_n\}_{n=0}^N)<2\epsilon.$$ This implies that the family $$\{U_{2\epsilon}(x_n)\}_{n=0}^N$$ covers $T$, contradicting the assumption that $T$ is totally bounded.

Step2. There is a Banach space embedding $l^\infty\hookrightarrow C_u(T)$.

By defining $f_n(x):=\max\left\{1-\frac{d(x,x_n)}{\epsilon},0\right\}$ for all $n\in\mathbb{N}$, there exist functions $\{f_n\}\subset C_u(T;[0,1])$ such that $$f_n(x_n)=1,\quad f_n|_{T\setminus U_\epsilon(x_n)}=0,\quad n\in\mathbb{N}.$$ Using these, we can define the following bounded linear operator: $$(a_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}a_nf_n$$ This operator preserves the norm, so it is also a Banach space embedding.

Step3. $l^\infty$ is not separable.

The set of indicator functions $\{\chi_I\}_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.

Separable spaces do not have a non-separable subspace, so this constitutes a contradiction.

If there are no serious mistakes, I would like to post the same answer on the original query on Mathematics Stach Exchange.

Thank you all so much.

What I realized from Mr. Ozawa's comment is as follows:

Step1. There exists an $\epsilon>0$ and a sequence $\{x_n\}\subset T$ such that the open balls $\{U_\epsilon(x_n)\}_{n\in\mathbb{N}}$ are disjoint.

This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup_{n=0}^NU_\epsilon(x_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup_{n=0}^NU_\epsilon(x_n)$, we have $$d(x,\{x_n\}_{n=0}^N)<2\epsilon.$$ This implies that the family $$\{U_{2\epsilon}(x_n)\}_{n=0}^N$$ covers $T$, contradicting the assumption that $T$ is totally bounded.

Step2. There is a Banach space embedding $l^\infty\hookrightarrow C_u(T)$.

By Urysohn's lemma, there exist functions $\{f_n\}\subset C_u(T;[0,1])$ such that $$f_n(x_n)=1,\quad f_n|_{T\setminus U_\epsilon(x_n)}=0,\quad n\in\mathbb{N}.$$ Using these, we can define the following bounded linear operator: $$(a_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}a_nf_n$$ This operator preserves the norm, so it is also a Banach space embedding.

Step3. $l^\infty$ is not separable.

The set of indicator functions $\{\chi_I\}_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.

Separable spaces do not have a non-separable subspace, so this constitutes a contradiction.

If there are no serious mistakes, I would like to post the same answer on the original query on Mathematics Stach Exchange.

Thank you all so much.

What I realized from Mr. Ozawa's comment is as follows:

Step1. There exists an $\epsilon>0$ and a sequence $\{x_n\}\subset T$ such that the open balls $\{U_\epsilon(x_n)\}_{n\in\mathbb{N}}$ are disjoint.

This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup_{n=0}^NU_\epsilon(x_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup_{n=0}^NU_\epsilon(x_n)$, we have $$d(x,\{x_n\}_{n=0}^N)<2\epsilon.$$ This implies that the family $$\{U_{2\epsilon}(x_n)\}_{n=0}^N$$ covers $T$, contradicting the assumption that $T$ is totally bounded.

Step2. There is a Banach space embedding $l^\infty\hookrightarrow C_u(T)$.

By defining $f_n(x):=\min\left\{1-\frac{d(x,x_n)}{\epsilon},0\right\}$ for all $n\in\mathbb{N}$, there exist functions $\{f_n\}\subset C_u(T;[0,1])$ such that $$f_n(x_n)=1,\quad f_n|_{T\setminus U_\epsilon(x_n)}=0,\quad n\in\mathbb{N}.$$ Using these, we can define the following bounded linear operator: $$(a_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}a_nf_n$$ This operator preserves the norm, so it is also a Banach space embedding.

Step3. $l^\infty$ is not separable.

The set of indicator functions $\{\chi_I\}_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.

Separable spaces do not have a non-separable subspace, so this constitutes a contradiction.

If there are no serious mistakes, I would like to post the same answer on the original query on Mathematics Stach Exchange.

Thank you all so much.

What I realized from Mr. Ozawa's comment is as follows:

Step1. There exists an $\epsilon>0$ and a sequence $\{x_n\}\subset T$ such that the open balls $\{U_\epsilon(x_n)\}_{n\in\mathbb{N}}$ are disjoint.

This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup_{n=0}^NU_\epsilon(x_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup_{n=0}^NU_\epsilon(x_n)$, we have $$d(x,\{x_n\}_{n=0}^N)<2\epsilon.$$ This implies that the family $$\{U_{2\epsilon}(x_n)\}_{n=0}^N$$ covers $T$, contradicting thee assumption that $T$ is totally bounded.

Step2. There is a Banach space embedding $l^\infty\hookrightarrow C_u(T)$.

By Urysohn's lemma, there exist functions $\{f_n\}\subset C_u(T;[0,1])$ such that $$f_n(x_n)=1,\quad f_n|_{T\setminus U_\epsilon(x_n)}=0,\quad n\in\mathbb{N}.$$ Using these, we can define the following bounded linear operator can be defined: $$(a_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}a_nf_n$$ This operator preserves the norm, so it is also a Banach space embedding.

Step3. $l^\infty$ is not separable. The set of indicator functions $\{\chi_I\}_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.

Separable spaces do not have an non-separable subspace, so this constitutes a contradiction.

If there are no serious mistakes, I would like to post the same answer on the original query on Mathematics Stach Exchange.

Thank all so much.

What I realized from Mr. Ozawa's comment is as follows:

Step1. There exists an $\epsilon>0$ and a sequence $\{x_n\}\subset T$ such that the open balls $\{U_\epsilon(x_n)\}_{n\in\mathbb{N}}$ are disjoint.

This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup_{n=0}^NU_\epsilon(x_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup_{n=0}^NU_\epsilon(x_n)$, we have $$d(x,\{x_n\}_{n=0}^N)<2\epsilon.$$ This implies that the family $$\{U_{2\epsilon}(x_n)\}_{n=0}^N$$ covers $T$, contradicting the assumption that $T$ is totally bounded.

Step2. There is a Banach space embedding $l^\infty\hookrightarrow C_u(T)$.

By Urysohn's lemma, there exist functions $\{f_n\}\subset C_u(T;[0,1])$ such that $$f_n(x_n)=1,\quad f_n|_{T\setminus U_\epsilon(x_n)}=0,\quad n\in\mathbb{N}.$$ Using these, we can define the following bounded linear operator: $$(a_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}a_nf_n$$ This operator preserves the norm, so it is also a Banach space embedding.

Step3. $l^\infty$ is not separable.

The set of indicator functions $\{\chi_I\}_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.

Separable spaces do not have a non-separable subspace, so this constitutes a contradiction.

If there are no serious mistakes, I would like to post the same answer on the original query on Mathematics Stach Exchange.

Thank you all so much.