**Proof**: Let X be a topological vector space and let x, y ∈, such that x ≠ y.

Set K = {x} and C = {y}

Then K is a compact and C is a closed and that K ∩ C = Φ. Hence there exist a neighborhood v of zero such that (K + V) ∩ (C + V) = Φ.

Since (K + V) and (C+V) are neighborhoods of x and y respectively.

Hence, every pair of distinct points of X can be separated by open sets.

Hence X is a Housdorff space. (Proved)