Let $X = \mathbb{A}^2_k$ with the free $\mathbb{Z}/2\mathbb{Z}$-action $(x,y) \mapsto (-x, -y)$, restricted to $\mathbb{A}^2 \setminus \{0\}$. The quotient algebraic space is reduced because $\mathbb{A}^2 \setminus \{0\}$ is reduced. The atlas is a reduced scheme, so the algebraic space inherits reducedness.

Let $U = \mathrm{Spec}(k[x]/(x^2))$ with the trivial equivalence relation. The resulting algebraic space is $U$ itself, which is non-reduced. More interestingly, one can form non-reduced algebraic spaces as quotients: let $V = \mathrm{Spec}(k[\epsilon]/(\epsilon^2))[t]$ with a free $\mathbb{Z}/2\mathbb{Z}$-action $t \mapsto -t$. The quotient is a non-reduced algebraic space.

Let $A$ be an abelian surface over $k$ (which is smooth, hence normal). Let $G \subset A$ be a finite étale subgroup. Then $A/G$ is an abelian variety, hence normal. More generally, any quotient of a normal scheme by a free étale group action is normal.

Consider the cuspidal cubic $C: y^2 = x^3$ in $\mathbb{A}^2_k$. Any algebraic space with atlas $C$ (for instance, a quotient of $C$ by a free action) will fail to be normal at the cusp point.

Every affine scheme $\mathrm{Spec}(A)$ is separated. More generally, any algebraic space that is affine over a separated scheme is separated.

Every quasi-projective scheme over a field is separated. Therefore, any algebraic space that admits a quasi-projective atlas and whose equivalence relation is a closed subscheme of $U \times U$ is separated.

The line with doubled origin is not separated: the diagonal image in $(\mathbb{A}^1 \sqcup \mathbb{A}^1) \times (\mathbb{A}^1 \sqcup \mathbb{A}^1)$ is not closed. More precisely, the closure of the diagonal contains additional points coming from the two origins.

Hironaka's example of a proper non-projective variety over $\mathbb{C}$ yields, after a suitable quotient, a proper algebraic space that is separated (properness implies separatedness for algebraic spaces over a field when the algebraic space is of finite type).

Any algebraic space of finite type over a noetherian scheme is quasi-compact and quasi-separated. In particular, all algebraic spaces appearing in classical algebraic geometry (over fields) with finite type hypotheses are quasi-compact and quasi-separated.

Let $U = \coprod_{n \in \mathbb{N}} \mathrm{Spec}(k)$ (a countably infinite disjoint union of points). With the trivial equivalence relation, this gives a non-quasi-compact algebraic space (it is just the scheme $U$, which is not quasi-compact).

Let $k$ be a field. Any algebraic space $X$ locally of finite type over $k$ is locally noetherian. If additionally $X$ is quasi-compact and quasi-separated, then $X$ is noetherian.

A noetherian algebraic space $X$ has a well-defined notion of dimension: $\dim(X) = \dim(U)$ for any étale atlas $U \to X$. Since étale morphisms preserve dimension, this is independent of the atlas.

Let $X$ be a smooth $k$-scheme and $G$ a finite group acting freely on $X$ with $|G|$ invertible in $k$. Then $X/G$ is a smooth algebraic space. If $k$ is perfect, smoothness and regularity coincide.

Let $C_1$ and $C_2$ be two smooth curves meeting transversally at a point $p$. Take $U = C_1 \sqcup C_2$ with the equivalence relation identifying the two copies of $p$. The resulting algebraic space (which is a scheme) is connected but reducible.

Let $A$ be an abelian variety (irreducible, connected) and $G$ a finite group acting freely. Then $A/G$ is an irreducible connected algebraic space. If $A/G$ is a scheme, it is an irreducible variety.

The atlas morphism $U \to X$ is étale by definition. A morphism $f: X \to Y$ is étale if for some (hence any) atlas $V \to Y$ and atlas $U' \to X \times_Y V$, the composite $U' \to V$ is étale.

The morphism $X \to \mathrm{Spec}(k)$ is proper if $X$ is of finite type, separated, and universally closed over $k$. Hironaka's proper non-projective algebraic space is an example of a proper algebraic space that is not a scheme.

If $U$ is a smooth $k$-scheme of dimension $d$ and $G$ acts freely on $U$, then $\dim(U/G) = d$. Étale morphisms preserve dimension, so the atlas and quotient have the same local dimension.

A zero-dimensional algebraic space over a field $k$ is necessarily a scheme: it is a disjoint union of spectra of fields. This follows from the fact that an étale equivalence relation on a zero-dimensional scheme is necessarily trivial (every étale morphism to a point is a disjoint union of isomorphisms).