Infimum of norm of Span of a closed subspace $Y$ and $x \notin Y$ is larger than zero

Question

Let $Y $ be a proper closed subspace of a normed space $X$.

Let $z \in X\setminus Y$.

Then

$ \inf \{\| z + y \| \space | \space y\in Y \}>0.$

Is this proposition true?

Answer 1

If $$ \inf_{y\in Y}\|z+y\|=0, $$ then there would exist a sequence $\{y_n\}_{n\in\mathbb N}\subset Y$, such that $$ \|z+y_n\|\to 0, $$ or equivalently, $y_n\to -z$, which would mean that $-z$ lies in $\overline{Y}$ the closure of $Y$. But $Y$ is closed, and hence $\overline{Y}=Y$. Hence $-z\in Y$ and thus $z\in Y$, since $Y$ is a subspace. Contradiction.

Answer 2

In a metric space for each subset , $d(x,Y)=0$ if and only if $x\in \overline{Y}$. So, being $Y$ a closed set, it contains each of its accumulation points. So, being the distance non-negative by definition and in this case different from $0$, you have the thesis.

Answer 3

$ \{\| z + y \| \space | \space y\in Y \} = \{d(z,-y) \space | \space y\in Y \} = \{d(z,y) \space | \space y\in Y \} $

Since $Y$ is closed and $z\notin Y$, $z$ is not a limit point of $Y$. Therefore $z$ is constantly bounded away from $Y$ by $\delta>0$.