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Under plus boundary conditions, if ${\sigma }_0=-1$, the negative cluster containing the origin is surrounded by a Peierls contour $\gamma$. Flipping all spins inside $\gamma$ is injective after $\gamma$ is specified and increases the Boltzmann weight by $e^{2\beta |\gamma |}$. ThereforeThe number of length-$m$ contours surrounding the origin is at most $Cm{3}^m$, so the sum tends to zero as $\beta \to \infty$, uniformly in $L$. For sufficiently large $\beta$ it is below $1/4$, and then

At zero field and with spin inversion symmetry in its boundary conditions, a finite Ising model pairs every configuration with its spin-reversed configuration. Their equal Boltzmann weights force the expected value of its spin magnetization $m=N^{-1}{\sum }_i{s}_i$ to vanish. Below a model's ordering critical temperature, an arbitrarily small positive field can instead select the positive spontaneous magnetization if size is taken to infinity first:The distinction is the order of limits and the stated symmetric boundary conditions; plus boundary conditions already break spin inversion symmetry. A zero-field symmetric mixture of the two ordered states can retain zero mean spin magnetization even in the thermodynamic limit. Also, $⟨|m|⟩>0$ in a finite system is not by itself proof of spontaneous symmetry breaking.^[ref]