Is a subspace of a vectorspace "closed"

Question

Suppose we have some vector space $V$ with subspace $U$. Is it true that if $v$ is a vector in $V$, and if $v+U$ is always in $U$, then $v$ is actually in $U$ itself?

The converse is obviously true, but I don't really know where to start with this.

Answer 1

The null vector is in $U$, and by hypothesis $v+0$ is in $U$, hence $v$ is in $U$.

Answer 2

Let me type what I think you're asking.

Suppose $V$ is a vector space, $U$ a subspace of $V$, and $v \in V$. If $$v + U = \{v + u : u \in U\}$$ is contained in $U$, then in fact $v \in U$.

Proof: Since $0 \in U$, we have $v + 0 \in v + U \subseteq U$, so there exists a $u \in U$ such that $v + 0 = u \in U$. But then $v = u \in U$. Q.E.D.

More generally we have a notion of cosets in a group. If $G$ is a group, $H$ is a subgroup of $G$, and $g \in G$, then we can talk about the set $gH = \{ gh: h \in H\}$. It is a fact that different cosets $g_1H, g_2H$ are either disjoint (their intersection is the empty set) or else they are equal.