Erwin Kreyszig's _Introductory Functional Analysis With Applications_: Section 2.3, Prob. 14 Here's problem 14 in the Problem Set immediately following Section 2.3 in the book, Introductory Functional Analysis With Applications by Erwin Kreyszig. 
Let $Y$ be a closed subspace of a normed space $(X, ||\cdot||)$. Show that a norm $||\cdot||_0$ on the quotient space  $X/Y$ can be defined as follows: 
$$||\hat{x}||_0 \colon= \inf_{x\in\hat{x}} ||x||$$ 
for all $\hat{x} \in X/Y$. 
Now my question is why do we require the subspace $Y$ to be closed? It is my feeling that it is to ensure that the norm on the quotient space satisfies the property
$$||\hat{x}||_0 = 0 \mbox{ if and only if } \hat{x} = \hat{0},$$ 
but I'm not completely sure. 
What if $Y$ is not a closed subspace? 
And, how to demonstrate all the properties of the norm in case $Y$ is a closed subspace? 
 A: Let us take a closer look at $\lVert\,\cdot\,\rVert_0$. Since for $x\in X$ we have $\hat{x} = \{ x-y : y \in Y\}$, we can write
$$\lVert \hat{x}\rVert_0 = \inf_{z\in\hat{x}} \lVert z\rVert = \inf \{ \lVert x-y\rVert : y \in Y\}.\tag{1}$$
So we have $\lVert \hat{x}\rVert_0 = 0$ if and only if for every $\varepsilon > 0$ there exists a $y_\varepsilon \in Y$ with $\lVert x-y_\varepsilon\rVert < \varepsilon$. We can write the latter as
$$\bigl(\forall\varepsilon > 0\bigr)\bigl(B_\varepsilon(x)\cap Y \neq\varnothing\bigr),\tag{2}$$
where $B_r(x) := \{ z\in X : \lVert x-z\rVert < r\}$ is the open ball with radius $r$ and centre $x$. But $(2)$ is just the criterion for $x \in \overline{Y}$, either by definition of, or by one of the first and most important criteria for, the closure of a set in a topological space.
The right hand side of $(1)$ is, for general subsets $Y\subset X$, the definition of the distance of the point $x$ to the subset $Y$, and we have
$$\overline{A} = \{ x \in S : \operatorname{dist}(x,A) = 0\}$$
for any subset $A$ of a metric space $S$.
Thus we have seen that
$$\{ x\in X : \lVert\hat{x}\rVert_0 = 0\} = \{ x\in X : \operatorname{dist}(x,Y) = 0\} = \overline{Y}.$$
But $\{ x\in X : \hat{x} = 0_{X/Y}\} = Y$, so if $Y$ is not closed, then there are elements $\xi$ of $X/Y$ with $\lVert\xi\rVert_0 = 0$ but $\xi \neq 0$, namely the elements $\xi = \hat{x}$ for $x\in \overline{Y}\setminus Y$.
Hence $\lVert\,\cdot\,\rVert_0$ can only be a norm on $X/Y$ if $Y$ is closed.
In all cases, however, $\lVert\,\cdot\,\rVert_0$ is a seminorm (inducing the quotient topology) on $X/Y$.
It is immediate that $\lVert\hat{x}\rVert_0 \geqslant 0$ for all $\hat{x}$, and that $\lVert\hat{0}\rVert_0 = 0$. The latter gives us the homogeneity
$$\lVert \alpha\cdot\hat{x}\rVert_0 = \lvert\alpha\rvert\cdot\lVert\hat{x}\rVert_0\tag{3}$$
for the case $\alpha = 0$, so we only need to prove it for $\alpha \in K\setminus \{0\}$, where $K$ is the scalar field (which needs to have a valuation $\lvert\,\cdot\,\rvert$ for $(3)$ to make sense; typically, $K$ is $\mathbb{C}$ or $\mathbb{R}$). For $\alpha\neq 0$, we have
\begin{align}
\lVert\alpha\hat{x}\rVert_0 = \lVert\widehat{\alpha x}\rVert_0 &= \inf \{ \lVert \alpha x-y\rVert : y\in Y\}\\
& = \inf \{ \lVert \alpha x - \alpha (\alpha^{-1}y)\rVert : y \in Y\}\\
&= \inf \{ \lVert \alpha (x-\alpha^{-1}y)\rVert : y \in Y\}\\
&= \inf \{ \lvert\alpha\cdot\lVert x-\alpha^{-1}y\rVert : y \in Y\}\\
&= \lvert\alpha\rvert\cdot \inf \{ \lVert x-\alpha^{-1}y\rVert : y\in Y\}\\
&= \lvert\alpha\rvert\cdot \inf \{ \lVert x- y'\rVert : y' \in Y\}\\
&= \lvert\alpha\rvert\cdot \lVert \hat{x}\rVert_0,
\end{align}
where in the penultimate line we used that $\alpha^{-1}Y = Y$. That establishes $(3)$ for nonzero $\alpha$, and together with the remark before, $(3)$ is established for all $\alpha \in K$ and $\hat{x} \in X/Y$.
It remains to check the triangle inequality for $\lVert\,\cdot\,\rVert_0$. Fix $x_1,x_2 \in X$ arbitrarily, and let $\varepsilon > 0$. By definition of $\inf$, there are $y_1,y_2 \in Y$ with
$$\lVert x_i - y_i\rVert < \lVert \hat{x}_i\rVert_0 + \frac{\varepsilon}{2}$$
for $i\in \{1,2\}$, and thus
\begin{align}
\lVert \hat{x}_1 + \hat{x}_2\rVert_0 &= \inf \{ \lVert x_1 + x_2 - y\rVert : y \in Y\}\\
&= \inf \left\{ \lVert (x_1 - y_1) + (x_2 - y_2) - (y-y_1-y_2)\rVert : y\in Y\right\}\\
&\leqslant \lVert (x_1-y_1) + (x_2 -y_2)\rVert\\
&\leqslant \lVert x_1 -y_1\rVert + \lVert x_2 - y_2\rVert\\
& < \lVert \hat{x}_1\rVert_0 + \frac{\varepsilon}{2} + \lVert\hat{x}_2\rVert_0 + \frac{\varepsilon}{2}\\
&= \lVert\hat{x}_1\rVert_0 + \lVert\hat{x}_2\rVert_0 + \varepsilon.
\end{align}
Since the inequality holds for every $\varepsilon > 0$, taking the infimum over the right hand side we conclude that indeed
$$\lVert\hat{x}_1 + \hat{x}_2\rVert_0 \leqslant \lVert \hat{x}_1\rVert_0 + \lVert\hat{x}_2\rVert_0.\tag{4}$$
Since $x_1,x_2$ were arbitrary, the triangle inequality $(4)$ for $\lVert\,\cdot\,\rVert_0$ is established.
We have seen above that $\lVert\hat{x}\rVert_0 = 0 \iff x\in \overline{Y}$, so the seminorm $\lVert\,\cdot\,\rVert_0$ is a norm on $X/Y$ if and only if $Y$ is closed.
It should be noted that in proving that $\lVert\,\cdot\,\rVert_0$ is a seminorm, only the seminorm properties of $\lVert\,\cdot\,\rVert$ were used, so we have the useful fact that every seminorm $p$ on $X$ induces a seminorm $p_0$ on the quotient $X/Y$, for every subspace $Y\subset X$, closed or not.
