Proving a set is compact and using density to show that something is not a homeomorphism

Question

Consider the following question

(1) Prove that if $X$ is a compact topological space, then a closed subset $Y$ of $X$ endowed with the subspace topology is compact. (2) Consider the following equivalence relation on $\mathbb{R}^2$ : $$ \left(x_1, y_1\right) \sim\left(x_2, y_2\right) \Longleftrightarrow\left(x_1-x_2, y_1-y_2\right) \in \mathbb{Z}^2 . $$ Let $X=\mathbb{R}^2 / \sim$ be the quotient space endowed with the quotient topology, and let $p: \mathbb{R}^2 \rightarrow X$ be the canonical surjection mapping each element to its equivalence class. Let $Z=\left\{(x, y) \in \mathbb{R}^2 \mid y=\sqrt{2} x\right\}$ (i) Show that $X$ is compact. (ii) Assuming that $p(Z)$ is dense in $X$, show that $\left.p\right|_Z: Z \rightarrow p(Z)$ is bijective but not homeomorphic.

(1) is easy enough- its theory.

I was able to do 2 (i) explicitly I think, however, there should be a better way, say be continuous maps. I thought of following: If I show that $f:\{ (x,y) : x^2+y^2 \leq 1\}\rightarrow X$ with $(x,y) \mapsto [(x,y)]$ is a continuous function, then I am done as the domain is compact by Heine-Bonel. This is easy enough as this a restriction to the projection map which is continuous. Is this correct?

For 2 (ii) showing infectivity is easy. However, I am struggling to see how one uses density to show that the given map is not e homomorphism.

I think we are suppose to show that $P(Z)$ is closed in $X$ and hence by part (1) we have $P(Z)$ compact. But $Z$ is not compact, so that can not be a homomorphism.

The issue is proving that $P(Z)$ is closed in $X$.

Question (s) :

1. Is my argument for 2 (i) correct?
2. How do I use density in 2 (ii)?

Answer 1

The second argument for 2(i) has the right idea, but the specific compact domain you chose won't surject onto $X$. Instead use something like $[0,1] \times [0, 1]$.

$p(Z)$ is in fact not closed in $X$. Otherwise, being dense, it would be all of $X$.

Here is a sketch. Consider $K = \{(x, y) \in Z \mid x \in [-1, 1]\}$ and $p(K) \subset p(Z)$. Then $K$ is compact and hence $p(K)$ is compact, hence closed. Now, $x = p((0, 0))$ is not an interior point of $p(K)$, for example $p((0, \epsilon))$ for $\epsilon > 0$ rational does not belong to $K$. So for any neighborhood $U$ of $p((0, 0))$ in $X$, $U \setminus p(K)$ is a non-empty open set of $X$ and hence contains some point of $p(Z)$, which must therefore be a point of $p(Z \setminus K)$. Now if $p|Z$ were to be a homeomorphism, then $p(\mathring{K})$ would be relatively open in $p(Z)$, ie be equal to $U \cap p(Z)$ for some $U$ open in $X$. But this is a contradiction, since $p((0, 0) )\in p(\mathring{K})$.