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Consider $l^{\infty}$ the vector space of limited sequences $(x_n)$ of real numbers with norm $||(x_n)|| = \sup |x_n|$. For $n \in \mathbb{N}$, let $e_n = (0,...,0,1,0,...)$ the sequence whose n-th coordinate is $1$. Let $F=\{(1+\frac{1}{n})e_n | n \in \mathbb{N} \}$. Show that $F$ is closed, $d(0,F) =1$, but $\nexists f \in F$ such that $d(0,f)=1$.

I tried solving this exercise, but I don't think I'm on the right way. Here is my attempt:

Since it is a normed space, it is $T_1$, then $x \in \bar{F} \iff \exists x_n \in F$ st. $x_n \to x$. Then for $F$ to be closed I need to find a sequence in F converging to a given point of $F$. Perhaps that's useful, but I can't think on a way to start a proof with this. Then I tried using this theorem: $x \in \bar{F} \iff d(x,F) = 0$

Let $(x_n) \in \bar{F}$.

$d((x_n),F) = \inf \{ d((x_n),(f_n)) | (f_n) \in F \} = \inf \{ ||(x_n)-(f_n)|| | (f_n) \in F \} = \inf \{ \sup|x_n-f_n|; (f_n) \in F \} = 0$

How can I conclude $(x_n) \in F$ ?.

$d(0,F) = \inf \{d(0,(f_n)) ; (f_n) \in F \} = \inf \{||(f_n)|| ; (f_n) \in F \} = \inf \{\sup |f_n|; (f_n) \in F \}$ . I can "see" that $n \to \infty$ there will be some $f_\infty$ whose non-zero coordinate will be $1$, then this value will be the infimum, but how can I write this?

2 Answers2

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The idea is $F$ is discrete. Every point in $F$ is at least 1 away from every other point. So any sequence in $F$ converging in $l^\infty$ has to eventually be constant.

You can find the distance to $F$ by computing the distance of each element to zero (which is just the infinity norm), then take the inf. $\inf 1+\frac{1}{n}=1.$

Finally, the distance of each element to zero is $1+\frac{1}{n}\neq 1$.

mathematician
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HINT: If $x\notin F$, there are two possibilities.

  • There are distinct $m,n\in\Bbb Z^+$ such that $x_m\ne 0\ne x_n$. Let $\epsilon=\min\{|x_m|,|x_n|\}$, and show that the open $\epsilon$-ball centred at $x$ is disjoint from $F$.
  • There is a unique $n\in\Bbb Z^+$ such that $x_n\ne 0$. Clearly $x_n\ne 1+\frac1n$; use this last inequality to find an $\epsilon>0$ such that the open $\epsilon$-ball centred at $x$ is disjoint from $F$.

For the rest, you know that

$$d(0,F)=\inf\left\{\left\|\left(1+\frac1n\right)e_n\right\|:n\in\Bbb Z^+\right\}\;;$$

and

$$\left\|\left(1+\frac1n\right)e_n\right\|=1+\frac1n\;,$$

so

$$d(0,F)=\inf\left\{1+\frac1n:n\in\Bbb Z^+\right\}=1\;.$$

However, for each $n\in\Bbb Z^+$ we have

$$d\left(0,\left(1+\frac1n\right)e_n\right)=1+\frac1n\ne 1\;.$$

Brian M. Scott
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  • $\epsilon = 1$? –  Oct 14 '16 at 21:06
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    @Alnitak: For the second case? No, that might be too big. If $x=e_n$, for instance, $d(x,F)=\frac1n$, which is less than $1$ for all $n\ge 2$. – Brian M. Scott Oct 14 '16 at 21:08
  • I think I'm having trouble calculating distances. Can you detail more? This is what I got for $x=e_n$: $d(x,F) = \inf { ||(1+\frac{1}{n})e_n - e_n|| | n \in \mathbb{N} } = \inf { \sup | \frac{1}{n} | } = \inf { \frac{1}{n} }=0 $ –  Oct 14 '16 at 21:17
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    @Alnitak: For $k\in\Bbb Z^+$ let $f_k=\left(1+\frac1k\right)e_k$. Then $d(e_n,f_n)=\frac1n$, and $d(e_n,f_k)\ge 1$ for $k\ne n$, since the $n$-th term of $e_n$ is $1$ and that of $f_k$ is $0$. Thus, $d(e_n,F)=\frac1n$. – Brian M. Scott Oct 14 '16 at 21:21
  • I don't get why in the first part $\epsilon = \min {|x_n|,|x_m|}$ works. Isn't $F = { 2e_1, (3/2)e_2, (4/3)e_3, (5/4)e_4,... }$, so when you take $x \notin F$, it could be something like $(0,42,43,...,0)$ and then $\epsilon = \min{42,43} = 42 $ which will be larger than $1$... –  Oct 14 '16 at 21:42
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    @Alnitak: If $d(y,x)<\min{|x_n|,|x_m|}$, then $y_m\ne 0\ne y_n$, so $y$ has at least two non-zero terms and is therefore not in $F$. – Brian M. Scott Oct 14 '16 at 21:46
  • would $\epsilon < \frac{1}{n}$ work in the second case? –  Oct 14 '16 at 21:58
  • @Alnitak: For the specific example of $x=e_n$, but not in general. You need to choose $\epsilon$ small enough so that if $d(y,x)<\epsilon$, then $y\notin F$. One way to do this is to choose it small enough so that if $|y_n-x_n|<\epsilon$, then $y_n\ne 1+\frac1n$. – Brian M. Scott Oct 14 '16 at 22:03
  • If $\epsilon = \frac{1}{|x_n|}$ then $y_n \neq 1+\frac{1}{n}$. Suppose by contradiction, then $\frac{1}{|x_n|} > |1+\frac{1}{n} - x_n| \Rightarrow \frac{1}{|x_n|} > |1+ \frac{1}{n}| - |x_n| \Rightarrow \frac{-1}{|x_n|} < -|1+\frac{1}{n}| + |x_n| \Rightarrow \frac{-1}{|x_n|} < |x_n| \Rightarrow -|x_n| > \frac{1}{|x_n|} \geq 0 $ Absurd. –  Oct 14 '16 at 22:29
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    @Alnitak: Your last implication is false. That choice of $\epsilon$ definitely won't work in general. I'll be mostly offline for a while, so I'm simply going to tell you: show that $\epsilon=\left|1+\frac{1}n-x_n\right|$ works. – Brian M. Scott Oct 14 '16 at 22:38
  • It is false indeed. Thanks for the replies and for your patience. –  Oct 15 '16 at 00:52
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    @Alnitak: You're welcome. – Brian M. Scott Oct 15 '16 at 06:14