Proving that $A$ is a closed set.

Question

The set A is defined as : $A := \left \{ x \in \mathbb{R}^d : \parallel x\parallel_{1} \leq 1 \right \}$. How can I go about proving that $A$ is a closed set.

I know that in this case, we are not using the Euclidean norm, because we have $\parallel x\parallel_{1}$ and not $\parallel x\parallel_{2}$. I know that the intersection of closed sets is closed.

I also know that a set is open if and only if it's equal to the union of a collection of open balls, so in this example, to show that the set is closed, the above can not hold.

I seem to be lost, on how to prove this problem, could anybody provide a proof for this problem ? Thank You!

Answer 1

Any norm is $1$-Lipschitz, hence continuous, as it satisfies the triangular inequality: $$\|x\|=\|x-y+y\|\leqslant\|x-y\|+\|y\|,$$ so that $\|x\|-\|y\|\leqslant\|x-y\|$, by symmetry, one finally gets $|\|x\|-\|y\||\leqslant\|x-y\|.$

Therefore, $A$ is a continuous preimage of a closed set, namely $[0,1]$, and so is closed.

Answer 2

The norm n(x) = ||x||, is a continuous function. A is the
inverse image of the closed set [0,1] by n, hence closed.