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Let $V$ be a vector space with a seminorm $\|\cdot\|_s$. Then apparently we can turn $\|\cdot\|_s$ into a norm $\|\cdot\|$ on $V/W$ by defining $\|v + W\| = \inf_{w \in W} \|v + w\|_s$ where $W$ is any closed subspace of $V$.

It's clear to me that if $V_0$ denotes the kernel of the seminorm $\|\cdot\|_s$ then $\|\cdot\|_s$ turns into a norm on $V/V_0$. What is not so intuitive to me is what happens if $W$ is disjoint from $V_0$. I think the fact that the norm $\|\cdot\|$ defined above then is a norm means that for every $v_0 \in V_0$ there is a sequence in $W$ converging to it. This holds because $0 \in W$ hence there is a sequence $w_n$ converging to $0$ and if $v_0 \in V_0$ then $v_0$ is in every neighbourhood of $0$, hence $w_n$ also converges to $v_0$. Is this correct?

If yes, would you show me some concrete examples illustrating this to help me develop some intuition? Thanks.

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You cannot «turn $\lVert\cdot\rVert_s$ into a norm» on the same vector space. It would be best to make your first paragraph be more precise. –  Mariano Suárez-Alvarez Jul 9 '12 at 7:00

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In general, we can turn $\|\cdot \|_s$ into a norm on $V/(W+ V_0)$ by the method described. To see this, note that if a norm is defined in this way on some quotient space $V'$ of $V$ we automatically get a surjection $V/W\to V'$ while in order for the norm to be a norm at all it the kernel must be $0$, so we get a surjection $V/V_0\to V'$. Putting these together gives us a surjection $V/(W+V_0)\to V'$, so all that remains is to verify $\|\cdot\|$ is a norm on $V/(W+V_0)$, which follows from composing the obvious maps $V\to V/V_0\to V/(W+V_0)$.

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