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Let $\mathbb{R}^{\infty}$ be the vector space of infinite sequences $(\alpha_1,\alpha_2,\alpha_3,\dots)$ of real numbers as a natural generalization of $\mathbb{R}^k$, $k\in\mathbb{N}^*$. Let $\boldsymbol{C}$ be the subset of $\mathbb{R}^{\infty}$ defined by, $$\boldsymbol{C} := \left\{\boldsymbol{p} = (p_1, p_2, p_3, \dots) \in \mathbb{R}^{\infty}\big/ \forall ~k, p_k \in [0, 1] \text{ and } \sum_{k = 1}^{\infty} p_{k} \leq 1\right\}.$$ I make these statements. Tell me if I am wrong.

  1. $\boldsymbol{C}$ is a compact and convex nonempty subset of the infinite dimensional space $\mathbb{R}^{\infty}$ for the the infinite norm defined by, $$||\boldsymbol{p}||_{\infty} = \max_k p_{k}.$$
  2. Schauder's fixed point Theorem (generalization of Brouwer's fixed point Theorem to an infinite dimensional space) can be applied: any continuous mapping from $\boldsymbol{C}$ to itself has a fixed point.
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  • $\begingroup$ It's not compact. Set $p^{(n)} = (\delta_{in})_{i \in \mathbb{N}}$. You will not be able to find a convergent subsequence. $\endgroup$ – Peter Morfe Jan 25 at 23:26
  • $\begingroup$ Thank you @PeterMorfe. I don't get it. What is the sequence you set? It seems that you have a sequence of sequence. $\endgroup$ – Ari.stat Jan 25 at 23:30
  • $\begingroup$ If $\sum_{i = 1}^{\infty}\delta_{in} = 1$ for all $n$ then the this is also true for the limit $(\bar{\delta}_i)$: $\sum_{i = 1}^{\infty}\bar{\delta}_{i} = 1$. $\endgroup$ – Ari.stat Jan 25 at 23:36
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Define a sequence $(p^{(n)})_{n \in \mathbb{N}}$ in $C$ by $p^{(n)} = (\delta_{in})_{i \in \mathbb{N}}$. In other words, \begin{equation*} p^{(n)}_{i} = \left\{ \begin{array}{r l} 1, & \text{if} \, \, i = n, \\ 0, & \text{otherwise.} \end{array} \right. \end{equation*}

$(p^{(n)})_{n \in \mathbb{N}}$ does not have any accumulation points in $C$ with the given norm topology. The idea is this: $\|p^{(n)} - p^{(k)}\|_{\infty} = 1$ for all $n,k \in \mathbb{N}$ so no subsequence is Cauchy.

Edit: Given $m \in \mathbb{N}$, if we define $C^{m}$ by \begin{equation*} C^{m}_{*} = \{x \in \mathbb{R}^{\infty} \, \mid \, \sum_{k = 1}^{\infty} |x_{k}| \leq 1, \, \, |x_{k}| \leq \frac{m}{k} \}, \end{equation*} then $C^{m}_{*}$ is compact in $\mathbb{R}^{\infty}$ with the norm $\|\cdot\|_{\infty}$ given above. Note that the set $C^{m}$ defined by Ari.stat below is simply $C^{m} = C^{m}_{*} \cap [0,1]^{\infty}$. Since $[0,1]^{\infty}$ is a closed subset of $\mathbb{R}^{\infty}$ with respect to $\|\cdot\|_{\infty}$, $C^{m}$ is a closed subset of $C^{m}_{*}$ and, thus, is itself compact.

To see this, first, notice that if $x \in C^{m}_{*}$, then $\|x\|_{\infty} \leq \sum_{k = 1}^{\infty} |x_{k}| \leq 1$.

To see that $C^{m}_{*}$ is compact, I will prove that any sequence in $C^{m}$ has a subsequence that converges. Suppose that $(x^{(n)})_{n \in \mathbb{N}} \subseteq C^{m}_{*}$. Since $|x^{(n)}_{k}| \leq 1$ independently of $n$ and $k$, we can employ a diagonalization argument to find a subsequence $(n_{j})_{j \in \mathbb{N}} \subseteq \mathbb{N}$ and a sequence $x \in \mathbb{R}^{\infty}$ so that, for each $k \in \mathbb{N}$, $\lim_{j \to \infty} x_{k}^{(n_{j})} = x_{k}$. In fact, we can show that $\lim_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} = 0$ must hold.

Indeed, given $\epsilon > 0$, if we choose $K_{0} \in \mathbb{N}$ such that $\frac{m}{k} \leq \epsilon/2$ for each $k \geq K_{0}$, then the inequality $\max\{|x^{(n)}_{k}|, |x_{k}|\} \leq \frac{m}{k}$ implies \begin{equation*} \|x^{(n_{j})} - x\|_{\infty} \leq \max \{ |x^{(n_{j})}_{1} - x_{1}|, \dots, |x^{(n_{j})}_{K_{0}} - x_{K_{0}}|, \epsilon \}. \end{equation*} Therefore, by pointwise convergence, $\limsup_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} \leq \epsilon$. We conclude by sending $\epsilon \to 0^{+}$.

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  • $\begingroup$ Thank you. So what is the contradiction here with the compactness? I understand that $(p^{(n)})_{n\in\mathbb{N}}$ does not have a limit in $C$. But anyway, $(p^{(n)})_{n\in\mathbb{N}}$ does not have a limit in $\mathbb{R}^{\infty}$. $\endgroup$ – Ari.stat Jan 26 at 0:22
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    $\begingroup$ If $C$ were compact, then any infinite subset would have to have at least one accumulation point. (This is true whenever we are working with compact sets in metric spaces.) $\endgroup$ – Peter Morfe Jan 26 at 0:46
  • $\begingroup$ Woww I got your point. Very useful. What if I defined the space as $$\boldsymbol{C^m} := \left\{\boldsymbol{p} = (p_1, p_2, p_3, \dots) \in \mathbb{R}^{\infty}\big/ \forall ~k, p_k \in [0, 1]; ~ p_k \leq \frac{m}{k}; \text{ and } \sum_{k = 1}^{\infty} p_{k} \leq 1\right\}?$$ I do not want to change my post because your answer is useful. Can you prove in another answer if the statements are true for some $m > 0$ with this new space? $\endgroup$ – Ari.stat Jan 26 at 13:52
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    $\begingroup$ @Ari.stat, I edited the answer to address the question in your last comment. $\endgroup$ – Peter Morfe Jan 30 at 19:36

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