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1. Let X be a topological vector space. , if A is a closed subset of X and x is an element of X that is not in A, then there exist disjoint open sets U1 and U2 such that x ∈ U1 and A ⊆ U2 . (2) X is connected. (3) X is compact if and only if X is {0}. (4) Every finite dimensional subspace Y of X is a closed subset of X. TOPOLOGICAL VECTOR SPACES 45 (5) If T is a linear transformation of X into another topological vector space X , then T is continuous at each point of X if and only if T is continuous at the point 0 ∈ X.

If z ∈ Z1 and z ∈ Z2 , then h(z ) ≥ −h(z). Indeed, z + z = z − v + z + v ∈ P. So, h(z + z ) = h(z) + h(z ) ≥ 0, and h(z ) ≥ −h(z), as claimed. Hence, we see that the set B of numbers {h(z )} for which z ∈ Z2 is bounded below. In fact, any number of the form −h(z) for z ∈ Z1 is a lower bound for B. We write b = inf B. Similarly, the set A of numbers {−h(z)} for which z ∈ Z1 is bounded above, and we write a = sup A. Moreover, we see that a ≤ b. Note that if z ∈ Z1 , then h(z) ≥ −a. Choose any c for which a ≤ c ≤ b, and define h on Z by h (z + tv) = h(z) − tc.

Therefore, X is pathwise connected, hence connected, proving part 2. Part 3 is left to an exercise. We prove part 4 by induction on the dimension of the subspace Y. Although the assertion in part 4 seems simple enough, it is surprisingly difficult to prove. First, if Y has dimension 1, let y = 0 ∈ Y be a basis for Y. If {tα y} is a net in Y that converges to an element x ∈ X, then the net {tα } must be eventually bounded in R (or C), in the sense that there must exist an index α0 and a constant M such that |tα | ≤ M for all α ≥ α0 .