The definition of noetherian scheme I have is that it can be covered by a fintie number of affine open subsets, where each of the affine open subset is the spectrum of a noetherian ring. I was curious is this equivalent (or implies?) to the topological space of the scheme being a noetherian topological space? Thank you.
On the definition of noetherian scheme
algebraic-geometrydefinitionschemes
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Disclaimer: I'm quite new to schemes so I might be making a silly mistake somewhere, but I think the answer to both questions is yes.
For Question 1 note that, by (2) and (3), the topology on $X$ is discrete, and hence in particular every point of $X$ is open. Since every open subset of $X$ has a cover by affine subschemes, in particular every singleton subset of $X$ is an open affine subscheme. So if the underlying points of $X$ are $p_1,\dots,p_n$, then the sets $U_i=\{p_i\}$ give an open cover of $X$ by affine subschemes; say $U_i$ is the spectrum of a ring $A_i$. By your observations each $A_i$ is local artinian. Moreover, since the $U_i$ are pairwise disjoint, by the sheaf property we have $\mathcal{O}_X(\bigcup_{i\in I}U_i)\cong\prod_{i\in I}A_i$ for every $I\subseteq[n]$, with the restriction maps being projections. So $X$ has the same underlying topological space and structure sheaf as $\operatorname{Spec}(A_1\times\dots\times A_n)$, ie is isomorphic to $\operatorname{Spec}(A_1\times\dots\times A_n)$. But $A_1\times\dots\times A_n$ is a product of artinian rings and hence artinian.
For Question 2 note first that $0$-dimensionality is inherited by subspaces. In particular, every open affine subscheme of $X$ has dimension $0$, and so by noetherianity every open affine subscheme of $X$ is the spectrum of an artinian ring. But every artinian ring is a product of finitely many local artinian rings, so that the spectrum of any artinian ring has the discrete topology, and hence $X$ has the discrete topology too. In particular, every point of $X$ is closed, and since $X$ is quasi-compact the underlying space of $X$ must be finite. So we are in the situation of Question 1 and hence done.
Fact: Let $X$ be a scheme and $U\subseteq X$ an open subset. Then $U$ (as a locally ringed space equipped with the restriction $\mathcal{O}_X|_{U}$) is a scheme.
Proof: Let $p\in U$ be a point. Since $X$ is a scheme, there is an open affine subscheme $V\subseteq X$ with $p\in V$; say $V\cong\operatorname{Spec}(B)$. Now $U\cap V$ is an open subset of $V$, and $p\in U\cap V$, so there is a basic open subset $W\cong D(b)$ of $V$, where $b\in B$, such that $p\in W\subseteq U\cap V$. But now $W$ is affine (it is isomorphic to $\operatorname{Spec}(B_b)$), and is an open neighborhood of $p$ in $U$. Since $p$ was arbitrary we conclude that $U$ is a scheme. $\blacksquare$
Best Answer
Noetherian schemes are always noetherian as a topological space, but the converse is false in general. To see that, pick some non-noetherian valuation ring (see here for examples) and consider the corresponding affine scheme.
Why does the implication hold? If a space $X$ is the union of finitely many noetherian subspaces $X_1,\dots,X_r$, then the space $X$ is noetherian itself and you can cover your noetherian scheme by finitely many noetherian spaces as the spectrum of a noetherian ring is noetherian.