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For any topological target $Y$, $f(K)$ is compact. Pull an open cover of $f(K)$ back by $f$, take a finite subcover of $K$, and map the covering statement forward. Without a separation assumption on $Y$, the image need not be closed, as a constant map into a nonclosed singleton of an indiscrete space shows.

If $Y$ is a metric space, $f(K)$ is also closed in $Y$, bounded, totally bounded, complete with its inherited metric, and sequentially compact. Closedness uses the Hausdorff theorem from part (iii); total boundedness follows by taking finite subcovers of the covers by balls of any fixed radius. Compact metric spaces are sequentially compact, and a Cauchy sequence with a convergent subsequence converges to the same limit, proving completeness. These properties are asserted for the image subset, without requiring the whole ambient metric space to be complete.

A metric space is complete if every Cauchy sequence in it converges to a point of that same space. A limit in a larger ambient completion is insufficient for this definition.

A sequence $(x_n)$ in a metric space is a Cauchy sequence ifThis compares sufficiently late terms with each other; it does not assume the existence of a limit in the space.

Yes. Continuity and nonvanishing on the compact interval give constantsConsequentlyA $d_{\alpha }$-Cauchy sequence is therefore supremum-norm Cauchy and has a continuous uniform limit $f$ by part (i). The upper bound then makes it converge to $f$ also in $d_{\alpha }$. Equivalently, multiplication by $\alpha$ is an isometry from this weighted space onto ordinary $C([0,1])$, since division by the continuous nonvanishing $\alpha$ gives its inverse.

This is completeness of a continuous weighted uniform norm.

Yes. A Cauchy sequence in the supremum norm converges in the given complete space $B([0,1])$ to a bounded function $f$. Supremum-norm convergence is uniform convergence. The uniform limit of continuous functions is continuous: for a point $x_0$ and tolerance $ϵ$, choose a continuous $f_n$ within $ϵ/3$ of $f$ uniformly, then use its continuity at $x_0$ and the triangle inequality. Hence $f\in C([0,1])$. Thus $C([0,1])$ is a closed, complete subspace of $B([0,1])$.

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence

A metric space is complete when every Cauchy sequence in it converges to a point of that space. A sequence is Cauchy when, for each $ϵ>0$, there is $N$ such that $d(x_m,x_n)<ϵ$ whenever $m,n\ge N$. The requirement that the limit lie in the space distinguishes completeness from having limits only in a larger ambient space.

Show that the subspace $C$ of convergent real sequences is closed in $X$. Suppose $a^{(m)}\in C$ converges uniformly in the coordinate index to $a\in X$. Given $ϵ>0$, choose $m$ so ${\sup }_j|a_j-a_j^{(m)}|<ϵ/3$. Since the fixed sequence $a^{(m)}$ converges, it is Cauchy, so for sufficiently large $j,k$ its coordinate difference is below $ϵ/3$. ThenThus $a$ is a Cauchy sequence of real numbers and converges by completeness of $\mathbb{R}$. Hence $a\in C$, so $C$ is closed. Part (a)(iii) proves the convergent-sequence subspace is complete.

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence

A metric space is complete when every Cauchy sequence in it converges to a point of that space. A sequence is Cauchy when, for each $ϵ>0$, there is $N$ such that $d(x_m,x_n)<ϵ$ whenever $m,n\ge N$. The requirement that the limit lie in the space distinguishes completeness from having limits only in a larger ambient space.

Show that the subspace $C$ of convergent real sequences is closed in $X$. Suppose $a^{(m)}\in C$ converges uniformly in the coordinate index to $a\in X$. Given $ϵ>0$, choose $m$ so ${\sup }_j|a_j-a_j^{(m)}|<ϵ/3$. Since the fixed sequence $a^{(m)}$ converges, it is Cauchy, so for sufficiently large $j,k$ its coordinate difference is below $ϵ/3$. ThenThus $a$ is a Cauchy sequence of real numbers and converges by completeness of $\mathbb{R}$. Hence $a\in C$, so $C$ is closed. Part (a)(iii) proves the convergent-sequence subspace is complete.

Define $\boxed{\Vert v\Vert =\sqrt{⟨v,v⟩}}$. Positive definiteness gives nonnegativity and separation of zero, and bilinearity gives absolute homogeneity. Cauchy–Schwarz follows by minimizing the nonnegative quadratic $\Vert u-tv{\Vert }^2$ in real $t$. It then gives $\Vert u+v{\Vert }^2\le (\Vert u\Vert +\Vert v\Vert {)}^2$, establishing the triangle inequality. Thus this is a norm.

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.

A metric space is complete if every Cauchy sequence of its points converges to a point of that same space. The requirement on the location of the limit distinguishes, for example, $\mathbb{R}$ from its incomplete subspace $\mathbb{Q}$.

A sequence is Cauchy ifThe comparison is between sufficiently late terms, without presupposing a limiting point in the space.

Continuity and nonvanishing on the compact interval implyHence $m\Vert f-g{\Vert }_{\infty }\le d_{\alpha }(f,g)\le M\Vert f-g{\Vert }_{\infty }$. A $d_{\alpha }$-Cauchy sequence is therefore uniformly Cauchy, so (i) gives a continuous uniform limit; the upper bound then gives convergence in $d_{\alpha }$. Thus $\boxed{(C([0,1]),d_{\alpha })\text{ is complete}}$. The positive lower bound is what makes the argument work.

Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

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 $f(x_0)=0$.

Assume $a,b>0$, the standard deviations. Substitute the two normal densities into (ii):The last step uses ${\int }_0^{\infty }u{e}^{-A{u}^2/2}du=1/A$. With $c=b/a$ this is $\boxed{f_V(v)=c/[\pi (c^2+v^2)]}$, the centered Cauchy density of scale $c$.

The desired event is $|V|<1$, outside the null event $X=0$. Integrating the density gives

For $x\ne 0$, the envelope ratio isIf $|x|\le 1$, use $|\sin x|\le |x|$ to bound this by $\frac{3}{2}(1+x^2)\le 3$. If $|x|\ge 1$, use $|\sin x|\le 1$ to bound it by $\frac{3}{2}(x^{-4}+x^{-2})\le 3$. The boundary equalities in these bounds cannot both occur in the sine estimate, so $M=3$ is a valid envelope. At zero the continuous ratio is $3/2$.

Generate independent Cauchy proposals as in (i), together with independent uniform $U$. Accept a proposal $Y$ whenusing the limiting value $1/2$ at $Y=0$. Repeat until acceptance. Part (a) proves exactness; the overall acceptance rate is $1/3$ and the mean number of proposals is three. No numerical maximization of the ratio is needed for correctness.