Is this quotient topology compact?

Question

Let $X$ be punctured Euclidean 3D space (or what's the term for this?) $$ X = \mathbb{R}^3 \setminus \{(0, 0, 0)\} $$

with the Euclidean topology and let ∼ be the relation defined as follows: $$ (x_1,y_1,z_1) ∼ (x_2,y_2,z_2) \iff \exists \: \mathrm{n} ∈ \mathbb{Z} \: \mathrm{such} \: \mathrm{that} \:(x_2, y_2, z_2) = (2^nx_1, 3^ny_1, 6^{−n}z_1). $$

Let Y be the set Y = X/~.

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Is Y compact? I cannot figure out a way to proceed and I am not even sure of the answer. Before this, I have already proven that Y is connected, not Hausdorff and that the projection map is open and I don’t know if any of this might help in the proof. I know that X is not compact but that doesn’t mean anything I think. The only way I could think is by using the definition of compact space (each of its open covers has a finite subcover) but I don’t have a clear image of the topology on Y. Maybe a homeomorphism to a particular space? Maybe finding finite subcovers of of each open and saturared cover of X?

Answer 1

We simply define a continuous unbounded function $Y\to \Bbb R$, namely the map $(x,y,z)\mapsto xyz$ from $X$ factors through $Y$.

If $Y$ was compact, any continuous function from $Y$ would have compact image.