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Suppose a large deviation principle at speed $a_n$ has a good rate function with unique zero mu. At any slower speed $b_n\to \infty$ with $a_n/b_n\to \infty$, its rate is zero at mu and infinity elsewhere. On a closed set excluding mu, the positive original rate gives decay faster than every exponential at speed $b_n$. Neighborhoods of mu have probability tending to one, giving the slower-speed open-set lower bound. The new rate is good because its finite sublevel set is the singleton mu.

Suppose $Y$ is closed in a normal space $X$ and $|f|<1$ on $Y$ for a continuous real-valued $f$ on $X$. The closed set $Z=\{|f|\ge 1\}$ misses $Y$. Urysohn's lemma supplies $u:X\to [0,1]$ equal to zero on $Y$ and one on $Z$. Then $(1-u)f$ agrees with $f$ on $Y$ and has range in $(-1,1)$. This permits the inverse transform $s\mapsto s/(1-|s|)$ in the proof of unbounded Tietze extension.

A subsequential limit of a sequence $(a_n)$ is a value $L$ approached by some subsequence $(a_{n_k})$ with strictly increasing indices $n_k$. For $a_n=(-1{)}^n(1+1/n)$, the subsequences at even integers and odd integers give subsequential limits one and minus one. A finite value appearing among only finitely many initial terms does not become a subsequential limit merely by being a member of the image of a function represented by the sequence.

For a bounded sequence of real numbers, its subsequential limits form a nonempty bounded set that is also a closed set. The Bolzano-Weierstrass theorem guarantees nonemptiness, and its extreme values are the limit inferior and limit superior. The sequence in sparse peaks can determine a limit superior has the two subsequential limits zero and one, despite almost all indices in the natural density sense carrying zero.

A T3 space is a regular topological space which is also a T1 space. Every metric space has this property: singletons are closed, and a sufficiently small ball about a point outside a closed set has closure avoiding that set. A second-countable T3 space is a metrizable space by Urysohn's metrization theorem.

Take the circle $S^1=\{(x,y)\in {\mathbb{R}}^2:x^2+y^2=1\}$ with its subspace topology. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two coordinate charts obtained by stereographic projection areBoth take values in the entire real line. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other coordinate chart includes that point. Hence $U\cup V=S^1$.^[ref]

To verify the homeomorphisms, solve the displayed equations for $(x,y)$. The inverse functions areEach formula lies on the circle, avoids its excluded point, and is continuous for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are continuous bijections with continuous inverse functions, as required for coordinate charts. Since the Euclidean plane is Hausdorff and second countable, the subset $S^1$ with its subspace topology is too. This checks all three conditions for a topological manifold, rather than just drawing a smooth-looking curve.

On the overlap $U\cap V$, both $x=\pm 1$ are excluded, so $y\ne 0$. Using $y^2=1-x^2$, the topological transition map isFor example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two connected components of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured real line. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its inverse function are smooth functions away from zero, so this topological atlas also supplies a smooth atlas.

Why can neither chart be enlarged to a single coordinate chart on the whole circle? The circle is a closed set and a bounded set in the Euclidean plane, hence compact by the Heine-Borel theorem. A global coordinate chart would give a homeomorphism to a nonempty open set $O\subseteq \mathbb{R}$. Its image would be compact and therefore closed in the real line. By connectedness of the real line, this forces $O=\mathbb{R}$, which is not compact. This is the one-dimensional instance of no compact manifold has a single Euclidean chart. Locally the circle is indistinguishable from the real line; globally compactness obstructs a single chart.

Every second-countable space that is also a T3 space is a metrizable space. First regularity of a topological space and the Lindelöf property yield a normal space: cover each of two disjoint closed sets by countably many open sets whose closures avoid the other, and subtract the first $n$ opposite closures from the $n$th neighborhood. The resulting unions are disjoint open neighborhoods. Urysohn's lemma then gives, for each pair of basic sets ${\overline{B}}_n\subseteq B_m$, a continuous function equal to one on the smaller closure and zero outside the larger set. There are countably many such pairs. The shrinking form of regularity of a topological space shows these functions generate the topology, and $T_1$ makes them separate points. Their coordinate map embeds the space into the countable cube, with compatible metric $d(x,y)={\sum }_j{2}^{-j}|f_j(x)-f_j(y)|$. The bounded tails prove compatibility with the product topology. Thus a countable base converts separation axioms into a numerical distance.

The zero set of a function $f$ is the preimage $f^{-1}(\{0\})$, consisting of all its zeros of a function. For a real- or complex-valued continuous function, it is a closed set, because the preimage of the closed singleton $\{0\}$ under a continuous function is closed. A nonzero holomorphic function on a connected open set has isolated zeros by the identity theorem for holomorphic functions.