Theorem: (x +v) ∩ A ≠ Φ if and only if x ⊂ A – V .

Proof: Suppose a ∈ (x + v) ∩ A. Then a ∈ x + V and a ∈ A.

But, a ∈ x + V

⟹ a = x + v

⟹ x = a – v ∈ A – V

⟹ x ∈ A – V

Conversely, if x ∈ A – V. Then

a = x + v ∈ X + V

⟹ a ∈ x + v

∴ a ∈ (A ∩ x +v)

∴ (x + v) ∩ A ≠ Φ (Proved).

(a) Let X be a topological vector space.If A ⊂ X then A = n (A + V), where V runs through all neighborhoods of O (zero).

Proof: Since V is a neighborhood of zero (0) if and only if x + v is a neighborhood of x. Hence x ∈ __A __if and only if (x + v) ∩ A ≠ Φ ∀ neighborhood V of 0 (zero) and this happens is x ∈ A – V for every such V.

Since – V is a neighborhood of 0 (zero) if and only iff V is a neighborhood of 0 (zero).

∴ x ∈ A + V for every neighborhood of 0 (zero).

Hence __A __= ∩ (A + V), where V runs through all neighborhood of zero.

(Proved)

(b) Let X be a topological vector space.If A ⊂ X and B ⊂ X, then __A__ + __B__ ⊂ __A + B__.

Proof: Let x ∈ __A__ + __B__. Then x = a + b, for some a ∈ __A__ and__ b__∈__ B__. Let W be a neighborhood of x. The continuity of addition operation implies ∃ neighborhoods W_{1} and W_{2} of a and b respectively such that

W_{1} + W_{2} ⊂ W.

Since a ∈ __A__, A ∩ W_{1} ≠ Φ similarly B ∩ W_{2} ≠ Φ.

Hence ∃ U and V such that U ∈ A ∩ W_{1} and V ∈B ∩ W_{2}.

Hence U + V ∈ A + B and U + V ∈ W_{1} + W_{2} ⊂ W.

Thus (A + B) ∩ W ≠ Φ.

Hence every neighborhood of x intersects A + B. Hence x ∈ __A + B__

∴ __A__ + __B__ ⊂ __A + B__. (Proved)

(c) Let X be a topological vector space.If Y is a subspace of X, so is __Y__.

Proof: Suppose α and β be scalar’s. Since multiplication by a nonzero scalar to a closed set is closed. α__Y__ closed if α ≠ o.

Since α__Y __ contains αY.

Hence __αY__ ⊂ α__Y __

Also x ∈ α__Y __

⟹ x = αY for some y ∈ __Y__

Thus U ∩ Y ≠ Φ for every neighborhood U of y.

But then (αu) ∩ (αy) ≠ Φ.

Hence αY ∈ __αY__ ⟹ x ∈ __αY__

Thus α__Y __ ⊂ __αY__ .

Thus we have α__Y __ = __αY__ for all non zero scalar α.

If α = 0, then α__Y __ = __αY__ holds trivially.

Since __A__ + __B__ ⊂ __A + B__ hence α__Y __ + β__Y __ = __αY__ +__βY __⊂ __αY +βY __⊂ __Y__.

The assumption that Y is a subspace was used in the last inclusion.

Hence __Y__ is a subspace of X. (Proved)

(d) Let X be a topological vector space.If C is a convex subset of X, so are __C__ and C^{0}.

Proof: Let 0 < t < 1. Since α__Y __ + β__Y __ = __αY__ +__βY__ hence we have

t__c__ + (1 – t )__c__ = __tc __+ __(1 – t )c__ ⊂ __tc + (1 – t )c__ ⊂ __C__.

As C is convex hence __C__ is convex.

Since C^{0} ⊂C and C is convex.

We have tc^{0} + (1 – t )c^{0} ⊂ tc + (1 – t )c ⊂ C.

Since tc^{0} and (1 – t)c^{0} are open sets, so it their sum is open. But c^{0} is largest open set contained in C.

Hence tc^{0} + (1 – t )c^{0} ⊂ c^{0}. Thus C^{0} is convex.

(e) Let X be a topological vector space. If B is a balanced subset of X, so is __B__; if also 0 ∈ B^{0} then B^{0} is balanced.

Proof: If 0 < | α | < 1 , then αB^{0} = (αB)^{0}, since x →αx is a

homeomorphism.Hence αB^{0}⊂αB ⊂ B, since B is balanced. But αB^{0} is open. So αB^{0} ⊂ B^{0}• If B^{0} contains the origin, then αB^{0} ⊂ B^{0} even for α =0. (Proved)

( f) Let X be a topological vector space. If E is a bounded subset of X, so is __E__.

Proof: Let V is a neighborhood of 0. Then by __U__⊂ W ∃ a neighborhood W of 0 such that __W__ ⊂ V. The boundness of E shows that ∃ a number s > o such that E ⊂ tW when t > s.

Hence __E__ ⊂ __tW__ when t>s ⟹ __E__ ⊂ t__W __⊂ tV when t>s.

Hence __E__ is bounded. (Proved).