For a topological vector space(tvs), I'd like to know whether
1.there exist a topological vector space V which is a Hausdorff space but does not satisfies the first countable axiom or
2.there exist a topological vector space V which is not a Hausdorff space and does not satisfies the first countable axiom.
I have questioned this problem but one may cannot see this. So I let it be alone and questioned this problem again.
I really find it difficult for me. Help me please. I can't work it out so far.
Please give me two examples about them. Thank you very much.
In a finite dimensional topological vector space $X$ consider the vector subspace $V:=\text{closure}(\{0\})$. You have $V = \bigcap \{U | U\text{ neighbourhood of }0\}$. This implies that the subspace topology of $V$ is the trivial topology, as every open set containing $0$ contains $V$.
Let $\{b_1,\dots,b_m\}$ be a vector space basis of $V$ and extend this by $\{b_{m+1},\dots,b_{m+n}\}$ to a basis of X. Now consider the vector space isomorphism $T:X\to V\times X/V$ defined by $$ T(\sum_{i=1}^{m+n} c_i b_i) = (\sum_{i=1}^m c_ib_i , (\sum_{i=m+1}^{m+n} c_ib_i) +V). $$ $T$ is continuous, as $\pi_V\circ T$ and $\pi_{X/V}\circ T = \pi_{X/V}$ are continuous. $T$ is open, as every open set containing $0$ contains $V$ and every surjective linear map onto a finite dimensional Hausdorff tvs is an open map [$X/V$ is a Hausdorff space, as the zero element $V$ is closed]. So altogether, $X\cong V\times \mathbb{K}^n$ where $V$ carries the trivial topology and $\mathbb{K}^n$ the euclidean topology.