Subspace of Noetherian Topological Space is Noetherian

Theorem
Let $X$ be a Noetherian topological space.

Let $Y \subseteq X$ be a subspace.

Then $Y$ is Noetherian.

Proof
Let $Y_1 \subset Y_2 \subset \ldots \subset$ be an ascending chain of open sets in $Y$.

By definition of subspace topology, there exists open sets $X_1, X_2, \ldots$ such that
 * $X_i \cap Y = Y_i$

for all $i$.

By taking the intersection $Z_i := \ds \bigcap_{j \mathop = i}^{\infty} X_j$, we have:


 * $Z_i \cap Y = Y_i$
 * $Z_1 \subset Z_2 \subset \dots$

Since $X$ is Noetherian, every ascending chain of open sets is eventually constant.

Hence $Z_i$ is eventually constant.

Then $Y_i = Z_i \cap Y$ is eventually constant.

Hence $Y$ is also Noetherian.

Also see

 * Space is Noetherian iff all Subsets are Compact