I am struggling with a subset of $\,l_\infty$ that has the following structure:
Let $a>0$ and consider set of sequences $S=\{x_k:k\in\mathbb{N}\}$ such that $x_{k,k}=a$ and $x_{k,i}=0$ for all $i\neq k$. (Here, $x_{k,i} $ is to denote the $i$-th element of sequence $k$)
Consider an arbitrary bounded sequence $y\in l_\infty$ that is not in $S$.
Since $y\neq x_k$ for all $k\in\mathbb{N}$, it is clear that the sup-metric $\,\sup_j|y_j-x_{k,j}|$ is strictly greater than zero. This, holds for all $k\in\,\mathbb{N}.$
I was wondering whether this is sufficient in order to be able to say that the infimum of $\sup_j|y_j-x_{k,j}|$ over $k\in\mathbb{N}$ is also strictly greater than zero for all bounded sequences $y$ that are not in $S$.
Intuitively, I could consider a sequence $y$ that has zeroes everywhere apart from its, say, 10th element which I choose to be $a+\varepsilon$ with $\varepsilon<a$ and arbitrarily small. Clearly, $\sup_j|y_j-x_{10,j}|=\varepsilon$ while $\sup_j|y_j-x_{k,j}|=a$ for all $k\neq 10.$
So, I kind of feel that this shows that $$\inf\,\{\,\sup_j|y_j-x_{k,j}|\,:\, k\in\mathbb{N}\,\}\,=\,\varepsilon\,>\,0$$
However, since $\varepsilon$ is arbitrarily close to zero, I am worried that the fact that the infimum is the greatest lower boundary implies that the infimum is 0. On the other hand, it seems that for any $y\not\in S$ every element in $\{\,\sup_j|y_j-x_{k,j}|\,:\, k\in\mathbb{N}\,\}$ must be strictly positive and the infimum of these elements is just the minimum which is thus strictly positive.
As you might have gleaned from my struggle, I am not extremely comfortable with the concept of an infimum.
Provided my intuition above is right, I would very much appreciate any pointers on how would I go about proving that $$\inf\,\{\,\sup_j|y_j-x_{k,j}|\,:\, k\in\mathbb{N}\,\}\,>\,0$$ for any bounded $y\not\in S$
Thank you very much.
The triangle inequality is what you need. Note that if $k \ne l$, then $\|x_k - x_l\| = |a|$. Therefore if $\|y - x_k\| < \frac {|a|}2$, then $\|y - x_l\| > \frac {|a|}2$.