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At parameter $1/2$, primal and dual bond percolation on the square lattice have the same probability distribution. Suppose an infinite percolation cluster exists. Large boxes meet it with probability tending to one. The square-root trick for positively associated events and quarter-turn symmetry imply that each side has an infinite exterior open arm with probability tending to one. The same holds for dual arms. A union bound then gives positive probability of primal top and bottom arms together with dual left and right arms.

Use a primal box $[-r,r{]}^2$ and a dual box with sides at $\pm (r+1/2)$. Open only primal edges whose endpoints both lie in the primal box, preserving all outward and exterior arm edges. The two primal arms are joined inside. Planarity prevents the two alternating dual arms from joining: a proposed dual connection together with the inner boundary arc separates one primal infinite arm from infinity. Thus the modified configuration has two infinite dual clusters. Its positive probability by finite modification of Bernoulli percolation contradicts uniqueness of the infinite percolation cluster. Consequently the percolation probability at $1/2$ is zero. This is often called Zhang's argument.

For bond percolation on the square lattice, put $g_n={\mathbb{P}}_p(0\leftrightarrow \partial [-n,n{]}^2)$. Splitting an open self-avoiding walk at its first visit to the radius-$m$ boundary gives disjoint certificates for reaching a boundary graph vertex and then reaching radius $n$ about that graph vertex. The van den Berg-Kesten inequality and the union bound give $g_{m+n}\le 8m{g}_m{g}_n$ for positive $m,n$. With $g_0=1$ and $|\partial {\Lambda }_0|=1$, the boundary-cardinality form holds for zero indices as well.

The Bonferroni correction tests each of $m$ hypotheses at level $\alpha /m$. The union bound then controls the familywise error rate by $\alpha$ under arbitrary dependence.

For arbitrary events $A_n$, if ${\sum }_n\mathbb{P}(A_n)<\infty$, then $\mathbb{P}({\mathrm{limsup}}_n{A}_n)=0$. The union bound gives $\mathbb{P}(\mathrm{limsup}{A}_n)\le {\sum }_{n\ge m}\mathbb{P}(A_n)\to 0$. No independence is required.

For the binomial random graph $G(n,1/2)$, the chromatic number is $(1+o(1))n/(2{\log }_{2}n)$ with high probability. A first moment method bounds the independence number above. Edge-disjoint clique packing in the complement graph, followed by an edge-exposure martingale and a union bound over moderately large vertex sets, supplies independent sets for greedy colouring by removing independent sets.

For independent bond percolation on a transitive locally finite lattice, let $c_n$ count rooted self-avoiding walks and let $\mu =\lim c_n^{1/n}$ be the connective constant. An infinite open percolation cluster supplies an open walk of every length. The union bound gives $\theta (p)\le c_n{p}^n$, which tends to zero when $p\mu <1$. Thus $p_c\ge 1/\mu$.

For any fixed density strictly below one, random subsets of an integer interval can have that density yet avoid arithmetic progressions longer than a constant times the logarithm of the interval length. A union bound over possible progressions bounds their occurrence, while concentration of the random cardinality supplies a dense realization. Passing to a subset gives an exact prescribed cardinality.

For a $d\times p$ matrix of independent standard normal entries and $0<t<1$, diagonal entries of $A^{T}A/d-I$ have two-sided tails bounded by $2e^{-d{t}^2/8}$, while off-diagonal entries have tails bounded by $2e^{-d{t}^2/4}$. A union bound over $p$ diagonal entries and $p(p-1)/2$ distinct off-diagonal entries controls their common entrywise maximum norm.

For nonempty index sets forming a laminar family of sets, reject an intersection hypothesis when the displayed adjusted value is at most $\alpha$. The inclusion-maximal true sets are disjoint and have total size at most $m$. Any false rejection forces some maximal true $J$ to have $p_J\le \alpha |J|/m$. Validity and the union bound control the familywise error rate by $\alpha$, without independence or a complete closure.

If an increasing event $A$ depends on $m$ sites or edges, the uniform-label monotone coupling of Bernoulli percolation and the union bound give $0\le {\mathbb{P}}_q(A)-{\mathbb{P}}_p(A)\le m(q-p)$ for $p\le q$. A change of the event requires a coordinate with label in $(p,q]$. For bond percolation from the origin to the boundary of $[-n,n{]}^2$, first-hit graph paths use the $4n(2n+1)$ internal edges, so the same count bounds the parameter sensitivity.

If the Holm step-down procedure makes a false rejection, it must reject the first true null in the ordered list. All $m_0$ true nulls remain at that step, so its cutoff is at most $\alpha /m_0$. The union bound on their marginally valid p-values gives familywise error rate at most $\alpha$. This proves strong control under arbitrary dependence, and super-uniformity suffices.

For the two-choice Achlioptas process, fix $a>0$ and $K\ge 1$. Conditional on any current graph with $N_{\ge K}\ge an$, put $W$ equal to the vertices in its large graph components. There are at most $n/K$ such initial graph components. If after $s$ more steps no graph component contains $an/3$ vertices of $W$, greedily assign final graph components to two groups so that each meets at least $|W|/3\ge an/3$ vertices of $W$. This induces a set partition of the initial large graph components, of which there are at most $2^{n/K}$ possibilities.

For any fixed such split $W=A\cup B$, a uniform candidate edge joins $A$ to $B$ with probability at least $2a^2/9$. If both candidates do so, the selected edge cannot avoid the crossing. Thus the probability that the split survives all $s$ steps is at most $(1-a^4/81{)}^s$. The deliberately weaker constant also covers distinct candidate sampling for sufficiently large $n$. In the absent-edge version, as long as the split has survived, every $A$--$B$ edge is absent, so conditioning on absence can only increase this crossing probability. A union bound with $s=\lceil A(a)n/K\rceil$ and $A(a)=81(\log 2+2)/a^4$ bounds failure by $e^{-2n/K}$. For fixed $K$, a further union bound makes the conclusion simultaneous over all starting steps $m\le 3n$. This proof is independent of how the rule chooses between the two candidates.

For $G$ with $n$ independent standard Gaussian rows in dimension $k$, combine a constant-radius sphere metric net, the quadratic form net bound, the chi-squared concentration inequality, and a union bound. This gives $P(\Vert G^{T}G/n-I_k{\Vert }_{\mathrm{op}}>1/2)\le 2\exp (k\log B-n/1024)$ for a numerical $B$. Taking $n$ larger than a constant times $k\log p$ gives exponential decay in $k\log p$. The subspace is fixed, so no union over supports is needed.

For independent standard normal random variables $Z_1,\dots ,Z_N$, the maximum of $|Z_j|$ has scale $\sqrt{\log N}$. A union bound and the Gaussian tail bound give the upper scale. Integrating the normal density over $[r,r+1/r]$ gives a lower tail bound $c{r}^{-1}{e}^{-r^2/2}$, implying $\mathbb{P}({\max }_j|Z_j|\le \sqrt{\log N})\to 0$.

Let $0<c<1$, $A>0$, $\sigma >0$ and $p\ge 2$. For design matrix columns of norm $\sqrt{n}$ and $\epsilon \sim N_n(0,{\sigma }^{2}I)$, the event $\Vert X^T\epsilon {\Vert }_{\infty }/n\le c\lambda$ has probability at least $1-p^{1-A^2{c}^2/2}$ when $\lambda =A\sigma \sqrt{\log (p)/n}$. Apply the sharp two-sided Gaussian tail bound to each standardized score and then the union bound; independence of the scores is not required.

For a fixed set $L$ of $\lfloor \sqrt{n}\rfloor$ vertices, let ${\tau }_L$ be the first step at which they all belong to one graph component. In the uniform random graph process, ${\tau }_L$ lies between $n(\log n-\omega )/4$ and $n(\log n+\omega )/4$ with high probability for every $\omega \to \infty$. Below that scale the marked isolated vertex count has divergent expected value and small relative variance. Above it the logarithmic-regime giant component misses $o(\sqrt{n})$ vertices; exchangeability and a union bound show that none are marked. The label coupling then transfers the two assertions to the process.

Suppose the one-arm probability on the cubic lattice satisfies ${\beta }_r=\exp (-\lambda r+o(r))$ with $\lambda >0$. Define $M_n$ as the largest maximum-norm radius of any cluster rooted in $[-n,n{]}^d$. Then $M_n/\log n\to d/\lambda$ convergence in probability. An upper bound is a union bound over $O(n^d)$ roots. For a lower bound, pack $\Theta ((n/\log n{)}^d)$ disjoint radius-$O(\log n)$ boxes and use the independent first-hit connection events inside them. Clusters are finite almost surely under the positive-rate hypothesis.

For independent Rademacher random variables and $\Vert u_i{\Vert }_2\le 1$, this bound holds for $\alpha >0$. The moment-generating function satisfies ${\prod }_j\cosh (t{u}_{ij})\le e^{t^2/2}$, with strict inequality for a nonzero vector and $t>0$. Applying the exponential Markov bound to both tails and then the union bound proves a strict failure bound; a zero vector has failure probability zero. At $\alpha =\sqrt{2\log (2m)}$, the success probability is strictly positive, ensuring a sign vector with all linear forms below that threshold.

Every circle rotation has the Furstenberg multiple recurrence theorem property for normalized Lebesgue measure. For a rational angle, use a period. For an irrational angle, choose a positive return time with angle close to zero; translation continuity in L1 on the circle then makes finitely many translates of a given positive-measure set simultaneously close to that set, and the union bound leaves a positive-measure intersection.