Proof of images of closed sets among continuous functions

Question

Let $g: \mathbb{R}^n \to \mathbb{R}^m$ with $g(x_1,...,x_n) := (e^{x_1}, 1,...,1)$ and the closed set $A := {x \in \mathbb{R}^n : -\infty < x_i \leq 0}$ for all $i = 1,...,n$} $\subset \mathbb{R}^n$

How can one find the image $g(A) :=$ {${g(x) : x \in A}$} $\subset \mathbb{R}^m$ of $A$ among $g$?

And how can one find out whether $g(A)$ is closed? Can I just take a subset $M \subset \mathbb{R}^m$ and say that it is closed if for every sequence $(x_k)_{k \in \mathbb{N}} \subset M$ with $x_k \to x \in \mathbb{R}^m$ (for $k \to \infty$) we can conclude, that $x \in M$?

Answer 1

You can find the image $g(A)$ by the following process:

1. By analysing the given function $g$, determine what values are plausible to be in the image. In this case, only the first coordinate varies, and so you can conclude that every point in the image will have its first coordinate of the form $e^x$, where $x\in A$.
2. Ask yourself: what are all the possible values that could possibly result by applying $e^x$ when $x=x_1$ and $x_1$ is the first coordinate of a point of $A$?
3. Answer your question from part 2. of the process.

This 3-step process should successfully lead you to solve the first part of your question. As for the second part, again, it is a 3-step process, as follows:

1. Remind yourself of what does it mean for a set to be closed. It seems that you already made an attempt in this direction. Make sure you know exactly what it means for a set in $\mathbb{R}^n$ to be closed.
2. Think about possible values you could put in the first coordinate of $g(x_1,\dots,x_n)$. In other words, think about what possible values could you generate by $e^x$, where $x$ is the first coordinate of a point in the set $A$. (Hint: let $x_1$ tend to $-\infty$).
3. Deduce from 2. that $g(A)$ is not closed.