# Show that the union $\{e_k \mid k \in \mathbb N\} \cup \{(1/k)_{k \in \mathbb N}\}$ is linearly independent on $\ell^\infty(\mathbb C)$

Let $$e_k$$ be the standard unit sequence, where all elements but the $$k$$th one are zeroes, and the remaining element is $$1$$. Show that the union $$$$\{e_k \mid k \in \mathbb N\} \cup \{(1 / k)_{k \in \mathbb N}\}$$$$ is linearly independent on $$\ell^\infty(\mathbb C)$$, where the sequence space $$$$\ell^\infty(\mathbb C) = \left\{ (x_k)_{k\in\mathbb N} \in \ell(\mathbb C) \mid \sup_{k\in\mathbb N} |x_k| < \infty \right\}\,.$$$$

An infinite set of vectors is said to be linearly independent, if the vectors of all its finite subspaces are linearly independent.

### Some thoughts

Looking at the set $$\{(1/k)_{k \in \mathbb N}\}$$ and the standard unit sequences $$e_k$$ it would seem that the harmonic sequence should be expressable in therms of these unit sequences. This is because multiplying $$e_k$$ by $$1/k$$ and then summing the resulting sequences together results in the harmonic sequence.

How could I use the fact that we are operating on $$\ell^\infty(\mathbb C)$$ to counteract this issue?

• The thing is that you would need an infinite sum to express the harmonic sequence as a linear expression of $e_i$. Linear dependence doesn't work over infinite sums. Jan 9 '20 at 19:24
• I don;t know where you got the problem from, but the notation is appalling: the set comprehension on the right of the union looks nothing like the singleton set containing the sequence $\langle 1/1, 1/2, 1/3, \ldots \rangle$. Jan 9 '20 at 21:13

Hint Assume that a finite linear combination of $$(e_k)$$ and $$s=(\frac{1}{n})$$ is zero. This means
$$c_1e_{k_1}+....+c_ne_{k_n}+cs=0$$ with $$c$$ potentially being $$0$$.
Look first at the "entry" $$m$$, where $$m > k_1,.., k_n$$ to deduce that $$c=0$$.
Next, look individually at each of the entries $$k_1,...,k_n$$.
• The first entry $m$ of the linear combination, as in $c(1/k)_{k \in \mathbb N}$, or the $m$th entry of the harmonic sequence itself? The latter makes more sense. Jan 9 '20 at 19:51
• @SeSodesa The entry of the linear combination. If $c_1e_{k_1}+....+c_ne_{k_n}+cs=0$ then the $m$th entry of $c_1e_{k_1}+....+c_ne_{k_n}+cs$ is zero. Jan 9 '20 at 20:22
• The point is that once $m>k_n,$ the $m$th component of the vector sum is $c/m$ and since each component is zero, $c=0.$ And then all the other coefficients are zero because the $\{e_k\}$ are linearly independent. Jan 10 '20 at 0:54