Solution. Non-negativity and symmetry are immediate, and \(d(x,y)=0\) exactly when \(x=y\) by definition. For the triangle inequality, if \(x=z\) the left side is \(0\) and there is nothing to prove; if \(x\neq z\) then \(y\) cannot equal both \(x\) and \(z\), so at least one of \(d(x,y)\), \(d(y,z)\) is \(1\) and the right side is at least \(1\).

The open balls are \[B(x;r) = \begin {cases}\{x\}, & r\leq 1,\\ X, & r>1.\end {cases}\] Since \(\{x\} = B\left (x;\tfrac 12\right )\) is open for every \(x\), and every subset is a union of its singletons, every subset of \(X\) is open. This is the discrete topology.

Solution. Let \(y\in B(x;r)\), so \(d(x,y)<r\). Put \(\delta = r - d(x,y) > 0\). If \(z\in B(y;\delta )\) then \[d(x,z) \leq d(x,y) + d(y,z) < d(x,y) + \delta = r ,\] so \(z\in B(x;r)\). Hence \(B(y;\delta )\subseteq B(x;r)\), and as \(y\) was arbitrary \(B(x;r)\) is open.

The proof is worth remembering as the model of its type: every openness argument in a metric space produces a \(\delta \) from the triangle inequality in exactly this way, and the choice \(\delta = r - d(x,y)\) is the natural one because it measures the room left.

Solution. Take \(U_n = \left (-\tfrac 1n,\ \tfrac 1n\right )\) for \(n=1,2,3,\dots \) Each is open, and \[\bigcap _{n=1}^{\infty }U_n = \{0\},\] since \(0\) lies in every \(U_n\) while any \(x\neq 0\) fails to lie in \(U_n\) once \(n>1/\left |x\right |\). The singleton \(\{0\}\) is not open in \(\mathbb {R}\), because no interval about \(0\) is contained in it.

There is no contradiction: the axioms permit arbitrary unions but only finite intersections. This example is precisely why the restriction to finitely many is there, and it is the mirror image of the closed-set example \(\bigcup _n\left [\tfrac 1n,1\right ] = (0,1]\).