Subspace of Noetherian Topological Space is Noetherian

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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 $Y$.

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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.

$\blacksquare$

Also see

• Space is Noetherian iff all Subsets are Compact