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$(V,T)$ topological vector space. U open neighbourhood of the origin $\Rightarrow$ $\exists N$ open neighbourhood of the origin s.t $\alpha N \subset U \ \forall |\alpha | \leq 1$

How could one prove that the above is true? I was thinking that maybe one can use that $U$ is absorbing. So for $\forall x \in N, \ \exists |\alpha| \leq 1 \in \mathbb{F}$ s.t $x \in \alpha U$ and then find some $\alpha$ which works $\forall x \in N$.

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    What's wrong with $N = \frac 1 2 U$? Did I misunderstand your question? – 2017-02-28
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    What if U is not bounded? Like, take a right half-plane in $\mathbb{R}$ and shift it slightly to the left. It is a neighbourhood of origin (open and contains origin). But you can't shrink it to be its proper set. – 2017-02-28
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    @Igor Why not? You might not shrink the part on the right going to infinity, but you shrunk the left side. – 2017-02-28
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    Right, my mistake... – 2017-02-28
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    @Olba12, Does your notation $\subset$ mean "any subset" or does it mean "proper subset"? – 2017-02-28
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    Yes... $U=\mathbb R$. – 2017-02-28
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    Maybe, you can take a neighbourhood with holes? – 2017-02-28
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    Take a union of a unit disk and annulus ($r_1 = 10, r_2 = 11$), both open. If you shrink it, wouldn't the annulus be displaced out of the original set. And still, the original set is open and contains the origin - hence it is a neighbourhood. – 2017-02-28

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If you take any vector space $V$ with the trivial topology $\{\varnothing, V\}$, you will find no such $N$. The only open neighborhood of any point is $V$ itself and you cannot scale it down to be a proper subset of $V$. If you dont talk about proper subsets, then $N=U$ is sufficient.

Note: I believe that this definition gives a topological vector space, because any map into such a topological space is continuous. So addition and scalar multiplication are continuous.

Note: Of course, this is no counter proof if we assume $T_1$ for such spaces. (thanks Igor)

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    From Wikipedia: "Some authors (e.g., Rudin) require the topology on X to be $T_1$. – 2017-02-28
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My suggestion would be to take a set $N$ of the form

$$N(U):=\frac12 \{x\in U\mid \alpha x\in U \text{ for all } |\alpha|\leq 1\}.$$

If $U=V$, then take $N(U')$ for some other open neighborhood $U'\subset V$ of the origin, which exists because of $T_1$. I think it is easy to see, that such an $N$ satisfies $\alpha N\subset U$ for all $|\alpha|\leq 1$. It remains to show that $N$ is open (it is non-empty, because $0\in N$).