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If a Galton-Watson process starts from one ancestor and has bounded nonnegative integer offspring $Z$ with $\mathbb{E}Z<1$, its total progeny $T$ obeys $\mathbb{P}(T>n)\le e^{-cn}$ for some $c>0$ and all integers $n\ge 1$. Explore one individual at a time. The unexplored queue has size $1+{\sum }_{j=1}^t(Z_j-1)$ until its first hit of zero, whose hitting time is $T$. Hence $T>n$ implies ${\sum }_{j=1}^n(Z_j-1)\ge 0$. Boundedness makes $M(\theta )=\mathbb{E}{e}^{\theta (Z-1)}$ finite near zero, with $M(0)=1$ and $M^{\prime }(0)=\mathbb{E}Z-1<0$. Choose $\theta >0$ with $M(\theta )<1$. The exponential Markov inequality gives the desired bound with $c=-\log M(\theta )$. The same proof works whenever the offspring law has an exponential moment near zero, even without boundedness.

The Heine-Borel theorem says that a subset $K$ of finite-dimensional Euclidean space is compact exactly when it is both closed and bounded:Here compactness refers to the subspace topology induced by the usual Euclidean metric, and closedness and boundedness are measured in the ambient Euclidean space. This characterization does not hold for arbitrary metric spaces.

The monotone convergence theorem states that if measurable functions $0\le f_n\uparrow f$ pointwise on a measure space, then $\int f_n\uparrow \int f$, allowing the limit to be infinite.

Extend each restriction to $I_n$ by zero on $I\setminus I_n$. The functions $f{1}_{I_n}$ are nonnegative and increase pointwise to $f$ because the intervals are increasing and exhaust $I$. HenceFor nonnegative $f$, integrability means finiteness of this integral, soNo boundedness or closedness of the intervals is required.

In any topological space a closed subset of a compact set is compact: add its open complement to an open cover and take a finite subcover of the compact set. However, compact subsets need not be closed in a general space. In the indiscrete two-point space a singleton is compact but not closed. Conversely a closed subset need not be compact: $\mathbb{R}$ itself is closed in $\mathbb{R}$ but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover.

In a Hausdorff space every compact subset $K$ is closed. Given $x\notin K$, separate $x$ and each $k\in K$ by disjoint neighborhoods; compactness selects finitely many neighborhoods covering $K$, and the intersection of their companion neighborhoods of $x$ misses $K$. Thus the complement is open. Closedness still need not imply compactness, as the same real-line example shows. It does imply compactness if the ambient space is compact.

In a metric space compactness implies boundedness: finitely many radius-one balls cover the set, so their centers have a finite maximal distance from any fixed center and the triangle inequality gives a global bound. The converse is false. An infinite set with discrete metric $d(x,y)=1$ for $x\ne y$ is bounded, but its singleton cover has no finite subcover. It is even closed and complete, so closedness plus boundedness alone does not suffice in general metric spaces.

The image $f(K)$ is compact: pull any open cover back by the continuous map, use a finite subcover of $K$, then map forward. For a general topological target, closedness is not guaranteed; a continuous image into the indiscrete two-point space can be a nonclosed singleton. No connectedness assertion follows merely from compactness of $K$.

If $Y$ is metric, $f(K)$ is compact, therefore closed in $Y$, bounded and totally bounded. It is also complete in its induced metric and sequentially compact. For the latter properties, compact metric spaces have a convergent subsequence of every sequence; a Cauchy sequence with one convergent subsequence converges to that same limit. Total boundedness follows by selecting finitely many balls of each prescribed radius. These are properties of the image itself, irrespective of whether $Y$ is complete.

Differentiability at $x_0$ means that a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists such that $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$. Since finite-dimensional linear maps are bounded, the right side tends to $f(x_0)$ as $h\to 0$, proving continuity.

For the specified scalar function, $0\le f(x,y)\le x^2\le x^2+y^2$. Dividing by $\sqrt{x^2+y^2}$ gives a bound tending to zero at the origin. Consequently $\boxed{df(0,0)=0}$ and the function is differentiable there.

For the final product assertion, write $f(x_0+h)=Lh+r(h)$, where $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,Cauchy–Schwarz and boundedness of $g$ near $x_0$ justify the last estimate. Thus $\boxed{dF(x_0)h=⟨df(x_0)h,g(x_0)⟩}$; differentiability of $g$ is unnecessary because $f(x_0)=0$.

Failure of boundedness: for every finite radius, some point lies outside the proposed bounding ball. A real-valued function is unbounded if its range is unbounded.