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

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

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 + BA + 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 W1 and W2 of a and b respectively such that

W1 + W2   ⊂  W.

Since a ∈ A, A ∩ W1 ≠ Φ similarly B ∩ W2 ≠ Φ.

Hence ∃ U and V such that U ∈ A ∩ W1 and V ∈B ∩ W2.

Hence U + V  ∈ A + B and U + V ∈ W1 + W2 ⊂  W.

Thus (A + B) ∩ W ≠ Φ.

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

A + BA + 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. αY closed if α ≠ o.

Since αY  contains αY.

Hence αY ⊂ αY

Also x ∈ αY

⟹ 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 αY  ⊂ αY  .

Thus we have  αY  = αY  for all non zero scalar α.

If α = 0, then αY  = αY  holds trivially.

Since A + BA + B hence αY  + βY  = 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 C0.

Proof: Let 0 < t < 1. Since αY  + βY  = αY +βY hence we have

tc + (1 – t )c = tc + (1 – t )ctc + (1 – t )cC.

As C is convex hence C is convex.

Since C0 ⊂C and C is convex.

We have tc0 + (1 – t )c0 ⊂ tc + (1 – t )c ⊂ C.

Since tc0 and (1 – t)c0 are open sets, so it their sum is open. But c0 is largest open set contained in C.

Hence tc0 + (1 – t )c0 ⊂ c0. Thus C0 is convex.

 

 

(e) Let X be a topological vector space. If B is a balanced subset of X, so is B; if also 0 ∈ B0 then B0 is balanced.

Proof: If 0 < | α | < 1 , then αB0 = (αB)0, since x →αx is a

homeomorphism.Hence αB0⊂αB ⊂ B, since B is balanced. But αB0 is open. So αB0 ⊂ B0• If B0 contains the origin, then αB0 ⊂ B0 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 EtW when t>s ⟹ E ⊂ tW ⊂ tV when t>s.

Hence E is bounded. (Proved).

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