Show that a set has no "unique nearest point" property

From Bridges' Foundations of Real and Abstract Analysis.

I'm given the following hint. Given ##a=(a_n)\in c_0\setminus S##, set ##\alpha=\sum _{n=1}^{\infty }2^{-n}a_n## and show that ##d(a,S)\leq|\alpha|##. Let ##x=(x_n)\in S##, suppose that ##\lVert a-x\rVert\leq|\alpha|##, and obtain a contradiction.

I can show that ##S## is a closed subspace of ##c_0## by considering the linear functional $$\phi : c_0 \to \Bbb{R} : (x_n) \mapsto \sum_{n=1}^\infty \frac{1}{2^n}x_n,$$ and show that it is bounded and hence its kernel is closed, which is ##S##. But I'm stuck at the second part. I do not really understand the hint. I do not know how to show ##d(a,S)\leq|\alpha|##, why we can suppose ##\lVert a-x\rVert\leq|\alpha|## and what would be contradicted. Grateful for any help.

What is [itex]d(a,S)[/itex]?

pasmith said:

What is [itex]d(a,S)[/itex]?

It is ##d(a,S)=\inf\{\lVert a-x\rVert :x\in S\}##, where the norm is ##\lVert a\rVert=\sup_n |a_n|##.

If the summation was [itex]\sum_{n=1}^\infty 2^{1-n}x_n[/itex], then we could say that [tex]
b_n = \begin{cases} a_1 - \alpha & n = 1 \\ a_n & n > 1 \end{cases}[/tex] is a member of [itex]S[/itex] which is a distance [itex]|\alpha|[/itex] from [itex]a[/itex]. Unfortunately because the summation is [itex]\sum_{n=1}^\infty 2^{-n}a_n[/itex], we have instead to set [tex]
b_n = \begin{cases} a_1 - 2\alpha & n = 1 \\ a_n & n > 1 \end{cases}[/tex] which is a distance [itex]2|\alpha|[/itex] from [itex]a[/itex].

If there exists an element [itex]x \in S[/itex] which is closest to [itex]a[/itex], then it satisfies [tex]0 < d(a,x) = d(a,S) \leq |\alpha|[/tex] (if the question is correct) because [itex]x \neq a[/itex]. So when looking for such an element it makes no sense to look for one with [itex]d(a,x) > |\alpha|[/itex].

pasmith said:

If there exists an element [itex]x \in S[/itex] which is closest to [itex]a[/itex], then it satisfies [tex]0 < d(a,x) = d(a,S) \leq |\alpha|[/tex] (if the question is correct) because [itex]x \neq a[/itex].

I don't understand your claim. Shouldn't it be ##0<d(a,x)=d(a,S)\leq 2|\alpha|##?

The following article was quite helpful (start reading from "A counterexample in the Banach space ##c_0##"). This treats a similar case as in the exercise above, except they use the summation ##\sum_{n=0}^\infty 2^{-n}x_n##. Does it make any difference if we use ##\sum_{n=1}^\infty 2^{-n}x_n## versus ##\sum_{n=0}^\infty 2^{-n}x_n##?

psie said:

The following article was quite helpful (start reading from "A counterexample in the Banach space ##c0##"). This treats a similar case as in the exercise above, except they use the summation ##\sum_{n=0}^\infty 2^{-n}x_n##. Does it make any difference if we use ##\sum_{n=1}^\infty 2^{-n}x_n## versus ##\sum_{n=0}^\infty 2^{-n}x_n##?

Certainly no difference in convergence in general, nor convergence to ##0##. A finite set of terms won't make a difference either way.

psie said:

I don't understand your claim. Shouldn't it be ##0<d(a,x)=d(a,S)\leq 2|\alpha|##?

By definition, [itex]d(a,x) = 0[/itex] if and only if [itex]a = x[/itex]. Here we know [itex]a \neq x[/itex] since [itex]x \in S[/itex] while [itex]a \notin S[/itex].

pasmith said:

By definition, [itex]d(a,x) = 0[/itex] if and only if [itex]a = x[/itex]. Here we know [itex]a \neq x[/itex] since [itex]x \in S[/itex] while [itex]a \notin S[/itex].

This explains why ##0<d(a,S)##, but not why ##d(a,S)\leq |\alpha|##. The example you gave with $$b_n = \begin{cases} a_1 - 2\alpha & n = 1 \\ a_n & n > 1, \end{cases}$$ shows that there is an element of ##S## with a distance ##2|\alpha|## from ##a##. So the distance ##d(a,S)## should be at most ##2|\alpha|##. What am I misinterpreting?

I suspect your ##d##is a distance function, but not a metric. Take, e.g., ##S:=\{(x,0): x \in \mathbb R\}## , i.e., the x-axis , and ##T:=\{(x,y): x,y\ in \mathbb R: xy=1\}##, i.e., the curve ##y=1/x; x>0## Then ##d(S,T)=0##, as the points in ##y=1/x## get indefinitely close to the x-axis as x goes to ##\infty##, so that ##d(S,T)=0##, but ##S\neq T##. Yes, the use of ##d## is confusing, maybe something like ##Dist(x,y)## would be better here.

psie said:

This explains why ##0<d(a,S)##, but not why ##d(a,S)\leq |\alpha|##. The example you gave with $$b_n = \begin{cases} a_1 - 2\alpha & n = 1 \\ a_n & n > 1, \end{cases}$$ shows that there is an element of ##S## with a distance ##2|\alpha|## from ##a##. So the distance ##d(a,S)## should be at most ##2|\alpha|##. What am I misinterpreting?

I am assuming that the question is correct in its assertion that [itex]d(a,S) \leq |\alpha|[/itex].

psie said:

I'm given the following hint. Given ##a=(a_n)\in c_0\setminus S##, set ##\alpha=\sum _{n=1}^{\infty }2^{-n}a_n## and show that ##d(a,S)\leq|\alpha|##. Let ##x=(x_n)\in S##, suppose that ##\lVert a-x\rVert\leq|\alpha|##, and obtain a contradiction.

Let ##a=(a_1,a_2,a_3,...)\in c_0\setminus S## be given and according to the metric

psie said:

It is ##d(a,S)=\inf\{\lVert a-x\rVert :x\in S\}##, where the norm is ##\lVert a\rVert=\sup_n |a_n|##.

We can check that ##\alpha## is finite and estimate range of values it can take
$$\left|\ \alpha \right| ~
= \left| \sum_{n=1}^{\infty} \frac{a_n}{2^n} \right|
\leq \sum_{n=1}^{\infty} \frac{\|a\|}{2^n}
= \|a\| \sum_{n=1}^{\infty} \frac{1}{2^n}
= \|a\|$$

Let ##x=(a_1-\alpha,a_2-\alpha,a_3-\alpha,...)## is given where
$$ \sum_{n=1}^{\infty} \frac{x_n}{2^n}
= \sum_{n=1}^{\infty} \frac{a_n-\alpha}{2^n}
= \sum_{n=1}^{\infty} \frac{a_n}{2^n} - \sum_{n=1}^{\infty} \frac{\alpha}{2^n}
=\alpha-\alpha
=0$$
and
$$d(a,x)
=\| a-x\|=\sup_{n \in \mathbb{N}} |a_n-x_n|
=\sup_{n \in \mathbb{N}} |a_n-a_n+\alpha|
= |\alpha|$$
but ( Edit: ##-\alpha## instead of ##\alpha## in lim ...)
$$\lim_{n \rightarrow +\infty} x_n = -\alpha \neq 0 \Rightarrow x \not\in S $$
Can you do something about ##x \not\in S## ?

Bosko said:

We can check that ##\alpha## is finite and estimate range of values it can take
$$\left|\ \alpha \right| ~
= \left| \sum_{n=1}^{\infty} \frac{a_n}{2^n} \right|
\leq \sum_{n=1}^{\infty} \frac{\|a\|}{2^n}
= \|a\| \sum_{n=1}^{\infty} \frac{1}{2^n}
= \|a\|$$

This computation shows ##d(a,S)\leq \lVert a\rVert##, but how does one show ##d(a,S)\leq |\alpha|##?

psie said:

This computation shows ##d(a,S)\leq \lVert a\rVert##, but how does one show ##d(a,S)\leq |\alpha|##?

TIP:
If ##\sum_{i=1}^{\infty} 1/2^i=1## and ##\sum_{i=n+1}^{\infty} 1/2^i=1/2^n##
how much is
$$\sum_{i=1}^{n} \frac{1}{2^i}=?$$

Can you make array of sequences
##x_1=(a_1-\text{"something"},a_2,a_3,...)##
##x_2=(a_1-\text{"something"},a_2-\text{"something"},a_3,...)##
....
##x_n=(a_1-\text{"something"},a_2-\text{"something"},a_3-\text{"something"},...,a_{n-1}-\text{"something"},a_n-\text{"something"},a_{n+1},...)##
...
so that they belong to ##S## ?

Check #2 post > https://www.physicsforums.com/threa...e-nearest-point-property.1059670/post-7055518
What @pasmith doing ?

Bosko said:

TIP:
If ##\sum_{i=1}^{\infty} 1/2^i=1## and ##\sum_{i=n+1}^{\infty} 1/2^i=1/2^n##
how much is
$$\sum_{i=1}^{n} \frac{1}{2^i}=?$$

From
$$\sum_{i=1}^{n} \frac{1}{2^i}=1-\frac{1}{2^n}=\frac{2^n-1}{2^n}$$
it follows
$$\frac{2^n}{2^n-1} \sum_{i=1}^{n} \frac{1}{2^i}=1$$
Multiplying both sides by ##-\alpha## we get
$$- \frac{2^n\alpha}{2^n-1} \sum_{i=1}^{n} \frac{1}{2^i}=-\alpha$$
For n=1 sequence ##x_1## is ##(a_1-2\alpha,a_2,a_3,a_4,...)## and the distance from ##a## to ##x_1## is ##2\alpha##.
For n=2 sequence ##x_2## is ##(a_1-4\alpha/3,a_2-4\alpha/3,a_3,a_4,...)## and the distance from ##a## to ##x_2## is ##4\alpha/3##.
For n=3 sequence ##x_3## is ##(a_1-8\alpha/7,a_2-8\alpha/7,a_3-8\alpha/7,a_4,...)## and the distance from ##a## to ##x_3## is ##8\alpha/7##.
...
The distance from ##a## to ##x_n## is ##2^n\alpha/(2^n-1)##