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I need some help in the follow question

The question is about the quotient norm:

"True or false: The infimum in the definition of the quotient norm: || [x]|| = inf m∈M ||x − m|| is always attained."

I saw this example in the internet but I don’t understand something:

the example was let $X=l^1$ and let $A=\{x \in l^1 | \sum_{n=1}^{\infty} {\frac{n}{n+1} x_n}=0 \}$ and we try to look at the unit vector $e_1$.

in the internet it clear that the $|| [e_1] ||_1 =0.5$ but I dont know why it is correct

thank to all

I get my example from this blog : http://calculus7.org/2012/03/11/nearest-point-projection-part-i-eschewing-transliterations/

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The author wrote

One can check that

the distance is $\frac12$. That doesn't (necessarily) mean it's obvious.

Let's see what we can find. For $a = 0$, we obviously have $\lVert e_1 - a\rVert_1 = 1$, so let's try to find something closer. Write

$$a = c_1\cdot e_1 - \sum_{k=2}^\infty c_k\cdot e_k,$$

where not all $c_k = 0$.

Then to have $a\in A$, we must have

$$\sum_{k=2}^\infty \frac{k}{k+1} c_k = \frac12 c_1,$$

and we have

$$\begin{align} \lVert e_1 - a\rVert_1 &= \lvert 1-c_1\rvert + \sum_{k=2}^\infty \lvert c_k\rvert\\ &> \lvert 1-c_1\rvert + \sum_{k=2}^\infty \frac{k}{k+1}\lvert c_k\rvert\tag{1}\\ &\geqslant \lvert 1-c_1\rvert + \left\lvert \sum_{k=2}^\infty \frac{k}{k+1}c_k\right\rvert\\ &= \lvert 1-c_1\rvert + \frac12\lvert c_1\rvert. \end{align}$$

Now it is easy to see that the last lower bound is minimised for $c_1 = 1$, with

$$\lVert e_1 -a\rVert_1 > \frac12.$$

So we need to see that we can come arbitrarily close to a distance of $\frac12$.

Well, for $a_k = e_1 - \dfrac{k+1}{2k}\cdot e_k,\; k \geqslant 2$, we have $a_k \in A$ and

$$\lVert e_1 - a_k\rVert_1 = \left\lVert \frac{k+1}{2k}e_k\right\rVert = \frac{k+1}{2k} = \frac12 + \frac{1}{2k} \to \frac12,$$

so indeed

$$\inf_{a\in A} \lVert e_1 - a\rVert_1 = \frac12,$$

and by $(1)$, the distance is not attained.

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