A topological vector space satisfies certain criteria, which will be presented below. As you might imagine, these criteria deal with open sets. There is no metric, no notion of distance, so that only leaves open and close sets and the operations of the vector space.
In a metric space, the translate of an open ball is an open ball, since the distance between two points does not change; but in a topological group, we have to use the properties of continuity to prove translation preserves open sets. If S is our topological space, S+S onto S is continuous. Let U be an open set and let V = U-x. Thus V is the preimage of U under translation. The preimage of U under addition is open, and is covered with base open sets. Remember that base open sets in the product are open sets cross open sets. For any y in V, x cross y is covered by an open set cross an open set. The open set that containes y has to lie in V, else its translate by x will not lie in U. Therefore V is covered by open sets and is open. Translation by x is continuous. Translation by -x is also continuous, hence translation is a homeomorphism on S.
When S is a topological module, a nearly identical proof shows that scaling by the units of the base ring is a homeomorphism. Of course our ring is a field, and every nonzero real number is a unit, so scaling by a nonzero constant carries open sets to open sets. Scaling by 0 is continuous, but certainly not a homeomorphism.
Let S be a topological group with a local base at 0. If an open set O contains x, O-x encloses 0, and contains an open set U from our local base. Thus O contains U+x contains x, and the translate of the local base at 0 gives a local base at x. This holds for all x, hence the translates of the local base at 0 give a base for the topology.
What's wrong with the discrete topology, where every set is open? Nothing, if S is a group, but watch what happens when S is a vector space. Consider the multiples of x by the reals in [0,1]. This is the smear of x by a closed interval. The set is open of course, and by continuity its preimage is open. Let y be a real number in [0,1] and x cross y is in an open set cross an open set, where the open set containing y lies in [0,1]. So [0,1] is covered by open sets and is open; yet it is closed. The topology of S is restricted by the topology of R.
The following properties turn a vector space into a topological vector space. They deal with the local base at 0, which is sufficient to describe the entire topology. The variables U V and W represent base open sets at 0.
Condition 1 builds a local base at 0, and condition 2 moves that local base to any point x, thus building a base for the topology. conversely, if S is a topological group, then the open sets containing 0 form a local base, and this or any other local base can be translated to any x.
Condition 3 implies, and is implied by, the continuity of S+S onto S. Assume the latter, and let U contain 0. Look at the preimage of S+S and find V+W in U. The intersection of these base open sets is another open set containing 0, which contains a base open set. Relabel this as V, and V+V is in U.
Conversely, assume condition 3, and consider the preimage of x+y+U. find V such that V+V lies in U. Now x+V + y+V lies in x+y+U, x cross y is contained in an open set, the preimage is open, and addition is continuous. A continuous group is equivalent to 1 2 and 3.
Assume scaling by R is continuous, hence continuous at 0. The preimage of the base open set U is open. An open interval about 0, cross an open set about x, lies in U. This means a nonzero constant times x lies in U.
A local base need not satisfy condition 5, but it's possible to find a new local base that does. Start with W, a set in the local base. continuity at 0 means an open interval (-e,e) times an open set V winds up in W. If e is 1 or greater, ratchet it down to a number below 1. Now c×V lies in W whenever |c| < e. Let U be the union of c×V for all c with |c| < e. Now U is an open subset of W. Replace W with U, and do the same for every other open set in the local base. Shrinking the open sets in a local base preserves the property of being a local base. Furthermore, each such set, multiplied by a constant below 1, produces a subset of the original. We are simply taking the union of fewer instances of c×V. Thus a topological vector space implies all 5 conditions.
Finally assume all 5 conditions hold. We already said 1 2 and 3 produce a topological group; we only need show scaling by R is continuous. Consider a base open set about cx, namely cx+U. Let V be a base open set such that V+V+V lies in U. Choose a real number d such that dx lies in V. If d > 1 let d = 1. (We can do this by property 5.) Multiply the interval (c-d,c+d) by the open set x+V/c, or by x+V if |c| < 1. The result lies in cx+U, and contains cx, hence multiplication by R is continuous.
A space need not be T0. Within R2, let vertical stripes, x = (-e,e), form a local base about 0. Verify that all 5 properties are satisfied. Yet points along the y axis cannot be separated.
If points are inseparable they can be clumped together to produce a quotient space. Does this disturb the structure of the group? If inseparable points become separated after translation, translate back and an open set contains one and not the other. Similar reasoning holds for multiplication. Thus the quotient space is also a quotient module, in this case a vector space. In the above example the kernel is the y axis, and the quotient space is the x axis, which is indeed a topological vector space.
Henceforth we will assume inseparable points have been clumped together, and our topological vector space is T1.
If x+U misses y, let V+V lie inside U, and suppose x+V and y-V intersect. Now x plus something in V yields y minus something in V, hence x plus something in U yields y, which is a contradiction. Therefore our topological vector space is T2, or hausdorff.
Let V be an open set about 0 in the range and pull back, by continuity, to U, an open set about 0 in the domain. Here V plays the role of ε and U plays the role of δ. Add x to the domain and x+U maps into f(x)+V. Given V, one U applies across the domain, and the function is uniformly continuous.