Theorem. An open set in real sequence space under the ℓ∞ norm is not totally bounded.

Proof. Consider an open set \(U\) containing a point \(p\). Suppose by way of contradiction that \(U\) is totally bounded. Then for every ε > 0, there exists a finite ε-net for \(U\). Fix ε, and let \(m\) be the number of points in our ε-net, which net we'll denote \(\{S_i\}_{i \in \{1, ..., m\}}\). We're going to construct a very special point \(y\), which does not live in \(U\). For all \(i \in \{1, ..., m\}\), we can choose the \(i\)th component \(y_i\) such that the absolute value of its difference from the \(i\)th component of the \(i\)th point in the net is strictly greater than ε (that is, \(|y_i - S_{i,i}| > \varepsilon\)) but also so that the absolute value of its difference from the \(i\)th component of \(p\) is less than or equal to ε (that is, \(|y_i - p_i| \le \varepsilon\)). Then for \(j > m\), set \(y_j = p_j\). Then \(|y - p| \le \varepsilon\), but that means there are points arbitrarily close to \(p\) which are not in \(U\), which is an absurd thing to happen to a point in an open set! But that's what I've been trying to tell you this entire time.