Image of a function can't contain an open set

Question

I am trying to prove that the image of $(x, y) \mapsto (x, xy, xye^{y})$ isn't a semianalytic set on $(0,0,0)$. In order to do so, I need to prove that its image can't contain an open set. How would I argue for this? It seems to me that because it is a function from a 2-dimensional manifold to 3-dimensional space, its image should at most be "2-dimensional". I know that there are pathological cases such as space filling curves, but I don't think that this would apply to the particular function chosen. Can anyone help me with this argument?

Answer 1

Your intuitive idea is correct, the image is indeed a 2D set.

To strictly prove this, you have several options. The easiest way is to notice that the curve $(0,t,0)$ can only ever intersect the image of the function in just one point.

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Or, even more strictly, if the image of the function contains an open set, then there must exist some neighborhood of $(0,0,0)$ that is contained in the image, and therefore, there must exist $\epsilon > 0$ such that the ball $B((0,0,0), \epsilon)$ is contained in the image of the set.

However, that would mean that $(0,\epsilon, 0)$ is in the image of the set, which is a contradiction.