Concrete counterexample for norms on quotient spaces

functional-analysislinear algebranormed-spacesreal-analysisvector-spaces

We have the following exercise in Functional Analysis:

Let $Y$ be a subspace of a normed linear space. Show that we can
define a norm on the quotient space $X/Y$ via

$$ \| [x] \|_q := \inf_{y \in [x]} \|y\|.$$

According to e.g. this answer, the statement should be wrong because $Y$ has to be closed which is not assumed in the exercise. However, the explanation given there is too short for me, so my question is: Could somebody provide a concrete counterexample which shows, that we have in general no positive definiteness if we not assume that $Y$ is closed?

Best Answer

Take$$X=\ell^\infty=\{\text{bounded sequences of real numbers}\}$$endowed with the norm $\bigl\lVert(x_n)_{n\in\mathbb N}\bigr\rVert_\infty=\sup_n\lvert x_n\rvert$. Let$$Y=\ell^1=\left\{(x_n)_{n\in\mathbb N}\in\ell^\infty\,\middle|\,\sum_{n=1}^\infty\lvert x_n\rvert<\infty\right\}.$$Finally, let$$x=\left(1,\frac12,\frac13,\frac14,\ldots\right),$$which is an element of $X$. Then $\inf\left\{\lVert x+y\rVert\,\middle|\,y\in Y\right\}=0$ because if$$y_n=\left(-1,-\frac12,-\frac13,\ldots,\frac1n,0,0,\ldots\right),$$then $y_n\in\ell^1$ and $\lVert x+y_n\rVert$ is as small as you wish.