If $A$ is open and contains $a$, there is an epsilon ball at $a$ contained in it. This ball is exactly $B_d(a,\epsilon )$ by the equality of balls proved above. Conversely such a ball supplies a neighborhood at $a$, and all other points of $A$ have their singleton neighborhoods inside $A$. Thus the stated ball condition is necessary and sufficient.

If $a\notin A$, every point of $A$ is isolated by (ii). The set $A$ is the union of these open singletons, so $\boxed{A\text{ is open in }(X,d_a)}$, including the empty case.

The same equality of balls gives the reverse implication: continuity at $a$ in $d$ implies continuity at $a$ in $d_a$. The latter implies continuity everywhere by the first part. These implications complete the equivalence of all three printed conditions.

For every $r>0$ one has $B_{d_a}(a,r)=B_d(a,r)$ because $d_a(a,x)=d(a,x)$. The epsilon–delta definition at $a$ is therefore identical for the two domain metrics. In particular continuity at $a$ in $d_a$ implies continuity there in $d$.

If continuity holds everywhere, it holds at $a$. Conversely, if it holds at $a$, it holds at every other point automatically: the singleton neighborhood from (ii) maps entirely to the single value at that point. Thus continuity everywhere in $d_a$ is equivalent to continuity at $a$ in $d_a$.

Use three nested sequences:For fixed $n,m$, letting $k\to \infty$ produces the point $(1/n+1/(n^{2}m),0)$. Letting $m\to \infty$ produces $(1/n,0)$, and letting $n\to \infty$ produces $(0,0)$. Accordingly, putThe successive derived sets areTo see that no limits are missing, take any convergent sequence of distinct points from $A_3$. If $n$ is unbounded, a subsequence has $n\to \infty$ and converges to the origin. Otherwise fix $n$ on a subsequence. Unbounded $m$ then forces limit $(1/n,0)$; with $n,m$ fixed, distinctness forces unbounded $k$, giving the corresponding point of $B$. The same argument, with one index fewer at each stage, proves the next two equalities. Distinct $n$-clusters are separated away from the origin because their first coordinates lie in the disjoint intervals $[1/n,1/n+1/n^2]$.

The original set has positive second coordinates, the first derived set has isolated $B$-points on the axis, the second has infinitely many $C$-points, and the third is a singleton. All three required successive inequalities follow. This is a three-stage accumulation tower.

TakeIts only accumulation point is $(0,0)$, which is absent from $A_2$. Every listed point is isolated, so none belongs to $A_2^{\prime }$. ConsequentlyFor completeness, away from the origin a sufficiently small disk meets only finitely many of the listed points and can be shrunk to exclude all but its centre, proving there are no further accumulation points.

Take $A_1={\mathbb{Q}}^2$. Every open disk contains infinitely many points with both coordinates rational, so every point of the plane is an accumulation point of $A_1$. Every disk also contains infinitely many points with an irrational first coordinate, and these lie in the complement. ThusThe infinitely-many-points observation ensures that one can avoid the centre of the disk itself in both cases.

If $A$ is open and contains $a$, it contains some $d_a$-ball about $a$, which is exactly the corresponding $d$-ball. Conversely, if $B_d(a,\epsilon )\subseteq A$, this is a $d_a$-neighbourhood of $a$, while every other point of $A$ has its open singleton inside $A$. Hence $A$ is open. Thus the stated ball-containment condition is necessary and sufficient.

Every point of $A$ differs from $a$, so its singleton is an open ball by part (ii). The union of these singleton open sets is $A$. Thus $A$ is open whenever $a\notin A$.

If $B\models T$ and $B\subseteq A^{\ast }\equiv A$, every universal sentence of $A$ holds in $A^{\ast }$ and is preserved by passage to the substructure $B$. Thus $B$ witnesses satisfiability of the joint theory.

For universal theory and substructures of elementarily equivalent models, conversely let $B$ satisfy $T$ together with the universal theory of $A$. Every existential sentence true in $B$ must be true in $A$: if an existential $\sigma$ were false in $A$, its universal negation would belong to ${\mathrm{Th}}_{\forall }(A)$ and hold in $B$, a contradiction. Hence ${\mathrm{Th}}_{\exists }(B)\subseteq \mathrm{Th}(A)$. In particular $\mathrm{Th}(A)\cup {\mathrm{Th}}_{\exists }(B)$ is satisfiable, with model $A$. Apply the preceding embedding result to the theory $\mathrm{Th}(A)$ and structure $B$. It supplies a structure $A^{\ast }\models \mathrm{Th}(A)$ containing $B$. Therefore $A^{\ast }\equiv A$ and $B\subseteq A^{\ast }$, as required. Elementary equivalence here does not assert that $A$ itself contains a copy of $B$.

If a model $B$ of $T$ contains $A$ as a substructure, every existential sentence true in $A$ remains true in $B$: use the same witness tuple and preservation of quantifier-free truth. Hence $B$ satisfies the required joint theory.

For the existential-theory embedding criterion, conversely suppose that joint theory has a model. Expand the language by a constant $c_a$ for each $a\in A$, and add the atomic and negated atomic diagram $D(A)$ to $T$. Any finite collection of its literals is a quantifier-free conjunction $\theta (\bar{c})$ true of the corresponding tuple in $A$. Replacing those named constants by variables and existentially closing gives a sentence in ${\mathrm{Th}}_{\exists }(A)$. A model of the assumed joint theory can supply its witnesses, and hence an expansion satisfying those finitely many diagram requirements and any finite subset of $T$.

Compactness therefore gives a model of $T+D(A)$. Its named Elements define an embedding of $A$ into its $L$-reduct: inequalities ensure injectivity, and all function and relation literals ensure preservation and reflection. Replace the ambient model by an isomorphic copy identifying this image with $A$ itself. Thus a model of $T$ containing $A$ exists. The diagram step explains how parameter-free existential information nevertheless yields an embedding of the whole structure.

No. With its induced usual order, the closed interval has a least element and a greatest element. At zero there is no smaller domain element, and at one there is no larger domain element. It fails the no-endpoint axioms, although it is densely ordered.

Yes. Removing zero leaves the induced order strict and linear, with no endpoints. For any $a<b$ there are infinitely many points in the real interval between them, so one can choose an intermediate point different from zero. More explicitly, the midpoint works unless it is zero, in which case $(2a+b)/3$ is nonzero and lies strictly between the endpoints. Thus density also survives.

The missing point is a cut, not a pair of adjacent elements; the axioms do not require all cuts to be filled.

Yes. The induced order is strict and linear and remains dense. For $0<x<1$, the points $x/2$ and $(x+1)/2$ lie in the domain and are strictly below and above $x$. Hence the open interval is a dense linear order without endpoints.

Yes. The usual order on the real numbers is strict and linear, midpoints witness density, and translation by $\pm 1$ supplies points on either side. It therefore satisfies the same dense linear order without endpoints axioms.

Yes. The usual order on the rational numbers is strict and linear. Between $a<b$ lies the rational midpoint $(a+b)/2$, and $a-1,a+1$ witness absence of endpoints. Thus it is a dense linear order without endpoints.

Every image of $L$ is a linear combination $a\cos 2x+b\sin 2x$. The function $\cos x$ is not such a combination. Indeed these functions are continuous, so equality almost everywhere would imply equality everywhere; evaluation at the endpoints would require simultaneously $a=1$ and $a=-1$. Therefore the forcing is outside the image and there is no solution. Orthogonality to the kernel is the equivalent image test for this self-adjoint finite-rank operator.

The forcing is $\frac{1}{2}\sin 2x$, which lies in the image of $L$. Since the eigenvalue on $\sin 2x$ is $\pi /4$, a particular solution is $y_p=(2/\pi )\sin 2x$. The full family isThus solutions exist and are not unique: adding any element of the infinite-dimensional kernel leaves $Ly$ unchanged. For example every constant can be added.