*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∈{1, ..., m}}. We're going to construct a very special point *y*, which does not live in *U*. For all *i* ∈ {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*| > ε) but also so that the absolute value of its difference from the

_{i,i}*i*th component of

*p*is less than or equal to ε (that is, |

*y*–

_{i}*p*| ≤ ε). Then for

_{i}*j*>

*m*, set

*y*=

_{j}*p*. Then |

_{j}*y*–

*p*| ≤ ε, 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.

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