Assume condition (1). For fixed $a$, the supremumis nondecreasing as $r$ increases. Therefore its limit as $r\downarrow 0$ exists in the extended real numbers and equals ${\inf }_{r>0}{M}_a(r)$. Because $a\in B(a,r)$, it is at least $f(a)$. Condition (1) says precisely that this limit is $f(a)$.

It follows that for every $\epsilon >0$ there is $r>0$ with $M_a(r)<f(a)+\epsilon$. Fix $C\in \mathbb{R}$ and $a$ with $f(a)<C$. Taking $\epsilon =C-f(a)$ produces an open ball on which $f(x)<C$. Thus $\{x:f(x)<C\}$ is an open set, and its complementThis proves (1) implies (2). The property being characterized is upper semicontinuity.

The answers areWrite $M_n(f,g)={\sup }_{0\le t\le n}|f(t)-g(t)|$. The given metric describes locally uniform convergence: $d(f_j,f)\to 0$ if and only if $M_n(f_j,f)\to 0$ for every fixed $n$. Indeed,so convergence in $d$ forces $M_n\to 0$. Conversely, split the defining series at $N$: its tail is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.

If $(f_j)$ is a Cauchy sequence in $d$, the same single-term estimate makes it a Cauchy sequence in the supremum norm on every $[0,n]$. The completeness of the continuous-map space in the uniform metric gives a continuous function $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the limit of a sequence gives $F_{n+1}{|}_{[0,n]}=F_n$, so these functions define a single continuous function $f$ on $[0,\infty )$. The preceding finite-head and tail argument proves $d(f_j,f)\to 0$. Thus the space is complete.

The constant functions $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no convergent subsequence, so the space is not sequentially compact.

For fixed $f,g$, set $\gamma (s)=(1-s)f+sg$. On $[0,n]$,as $s^{\prime }\to s$. The equivalence with locally uniform convergence proves that $\gamma$ is a continuous path for $d$. Hence the space is path-connected and connected.

The answers areDistinct natural numbers have $d(x,y)>1$. Taking $\epsilon =1$ in the Cauchy sequence condition forces every sufficiently late pair of terms to be equal. Every Cauchy sequence is therefore eventually constant and converges in $\mathbb{N}$, proving completeness.

The sequence $x_n=n$ has no convergent subsequence, since convergence would force its sufficiently late terms to lie within distance $1$ of its limit of a sequence, hence to equal that limit. Thus sequential compactness fails.

Every singleton is an open ball of radius $1$, so the metric topology is the discrete topology. For example, $\{1\}$ and $\mathbb{N}\setminus \{1\}$ form a separation into two nonempty open sets, so $X$ is not connected. It is consequently not path-connected either.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.

For $\alpha \ne 2$, both reduced forms are $I_3$, so $A^{\prime \prime }=(A^{\prime }{)}^T$. At $\alpha =2$, the reduced column echelon form is the transpose of the reduced row echelon form of $A^T$, givingFor example, $C_2\mapsto -C_2$ followed by $C_3\mapsto C_3+C_1-C_2$, using the updated second column, produces the displayed $A^{\prime \prime }$ directly. Its third column is zero. The two matrices differ, so (C) is true exactly when $\alpha \ne 2$. Taking a transpose exchanges the kind of reduction, but also replaces the original $A$ by $A^T$; it does not generally identify the two reduced forms of $A$ itself.

Every elementary column operation preserves the linear span of the columns: each new column is a linear combination of the old columns, and the inverse operation supplies the reverse inclusion. Thus the column space of $A^{\prime \prime }$ equals that of $A$, which is precisely the image of a linear map $T_{\alpha }$. Consequently (B) is true for every $\alpha \in \mathbb{R}$.

For $\alpha \ne 2$, $A^{\prime }=I_3$, whose row space is ${\mathbb{R}}^3=\mathrm{im}{T}_{\alpha }$. For $\alpha =2$, elementary row operations give the reduced row echelon formIts row space is $\{(u,v,w):u+v+w=0\}$, whereas the image of a linear map is $\{(u,v,w):u-v+w=0\}$. These differ: $(0,1,-1)$ belongs to the former but not the latter. Thus (A) is true exactly when $\alpha \ne 2$. Rows are being identified with vectors in ${\mathbb{R}}^3$ in this comparison.

A unitary matrix has $A^{\ast }A=I$. For example,This example is invertible, has nonreal off-diagonal entries, and preserves the standard Hermitian inner product. Although it is also symmetric, that additional property is not required for unitarity.

A Hermitian matrix has $A^{\ast }=A$. ChooseIts off-diagonal entries are complex conjugates and its diagonal entries are real, so the defining condition holds. It is also invertible.

A symmetric matrix has $A^T=A$, without conjugation. One invertible example with nonreal off-diagonal entries isIt is symmetric but not Hermitian, illustrating the importance of the distinction over $\mathbb{C}$.

For $c<0$, maximize $P$ to lower the quartic coefficient. The identity $P=[1-(n_x^4+n_y^4+n_z^4)]/2$ shows that its maximum $1/3$ occurs for $n_x^2=n_y^2=n_z^2=1/3$. ThereforeThere are eight symmetry-related easy directions. This is a cubic easy-axis anisotropy even though the selected directions are body diagonals; it is not a continuous easy plane. At $c=0$ the quartic free energy instead has a sphere of degenerate orientations to this order.

For $c>0$, minimize the nonnegative anisotropy invariant $P$. Its minimum is zero, achieved only when one squared component of the unit vector is one and the others vanish. Thus the easy directions areThe six signs and axes are symmetry-related ferromagnetic domains. Allowing the magnetization amplitude to relax does not change the result: its minimum free-energy reduction is proportional to $-1/(b+cP)$, which is most negative for the smallest positive quartic coefficient. A $⟨110⟩$ direction has $P=1/4$ and is not an equally stable easy direction.

A human during childhood. Developmental studies must specify their sampled age range rather than treating all children as one population; recruitment and task accessibility affect the scope of an inference.

A developmental period spanning changes in physical maturation, social roles and increasing autonomy. Clinical research should specify sampled ages and relevant developmental characteristics rather than treating an age label as a complete explanation of treatment response.

For traceless Hermitian matrices representing the Lie algebra of the special unitary group $SU(N)$, normalized by $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, completeness of $\{I/\sqrt{2N},T^a\}$ in the space of Hermitian matrices gives the displayed identity. In particular ${\sum }_a{T}^{a}X{T}^a=\frac{1}{2}\mathrm{tr}(X)I-X/(2N)$. For traceless $X=T^b$, this becomes ${\sum }_a{T}^a{T}^b{T}^a=-T^b/(2N)$. These identities reduce QCD color charge sums without choosing explicit matrices.

For the special unitary group $SU(N)$, put ${\epsilon }_i=e_i-N^{-1}{\sum }_j{e}_j$. Its defining representation has weight system $\{{\epsilon }_i\}$, and its exterior square has weight system $\{{\epsilon }_i+{\epsilon }_j:i<j\}$; each has multiplicity one. Their dual representations negate the weights of a representation. The Dynkin label coordinates are $({\epsilon }_i{)}_r={\delta }_{ir}-{\delta }_{i,r+1}$. These follow by evaluating diagonal traceless generators on coordinate vectors and their exterior products.

Under the standard embedding $SU(N-1)\subset SU(N)$, the last defining weight of a representation restricts to zero and the others become the defining weights of a representation of $SU(N-1)$. Splitting pairs according to whether they contain the last index proves the branching of a representation $V_N\downarrow =V_{N-1}\oplus 1$ and ${\Lambda }^2{V}_N\downarrow ={\Lambda }^2{V}_{N-1}\oplus V_{N-1}$, including multiplicities.

For ideal identical $R_x(\pi /2)$ pulses separated by $R_z(-\delta T)$, multiply the unitary operators to obtain $U=-i[\cos (\delta T/2)X-\sin (\delta T/2)Z]$. An input at either pole of the Bloch sphere returns to that same pole with probability ${\sin }^2(\delta T/2)$. In particular zero delay gives a full NOT, so its return probability is zero. Opposite-sign pulses instead give ${\cos }^2(\delta T/2)$. The distinction matters when identifying the measured quantum state population.

A geostationary orbit is a circular, prograde, equatorial orbit with period equal to one sidereal day, approximately $23\mathrm{h}56\min$. Its satellite therefore remains above one longitude on the equator. Merely having a one-day period gives a geosynchronous orbit; an inclined or eccentric version still moves across the ground observer's sky.

Balancing gravitational acceleration and centripetal acceleration givesOne route is injection into an elliptical geostationary transfer orbit, followed by a burn near apogee to circularize and remove the remaining inclination; direct or low-thrust transfers are also possible. Subsequent station keeping controls longitude, inclination and eccentricity.

The fixed wide view permits frequent observations of weather, clouds and rapidly changing storms, as in Meteosat and GOES. The tradeoffs are large range, generally coarser spatial sampling for comparable optics, oblique views at high latitudes and poor polar coverage. It is valuable for following change through the day rather than obtaining global coverage with one spacecraft.^[ref]

A Sun-synchronous orbit maintains approximately constant orientation relative to the mean Sun, so corresponding passes over a location occur at nearly the same local solar time. It is usually a nearly polar, slightly retrograde low Earth orbit. Earth rotates beneath the successive tracks, allowing broad coverage; the satellite is not stationary over a location and does not necessarily revisit that location every day.

It is attained by selecting altitude, orbital eccentricity and orbital inclination so that nodal precession caused by Earth's equatorial bulge matches the mean apparent annual motion of the Sun, about $360^{\circ }$ per year eastward. To leading order,Here $a$ is the semi-major axis, $e$ the orbital eccentricity, $i$ the orbital inclination, $n=\sqrt{G{M}_E/a^3}$ the mean motion, and $J_2$ Earth's dimensionless quadrupole coefficient. A retrograde inclination has $\cos i<0$, giving the required positive precession; typical imaging altitudes yield inclinations around $97^{\circ }$–$99^{\circ }$. The launch establishes the required orbital plane, and subsequent orbit maintenance limits drift.

This geometry suits repeated optical and multispectral imaging of land, vegetation and oceans, because comparable local time reduces illumination and shadow differences. Landsat and Sentinel-2 are examples. Radar missions such as Sentinel-1 also use this orbit for regular global sampling even though their illumination is supplied by the radar. Constant local time helps comparison but does not remove seasonal changes in solar elevation.^[ref]

Pore fluid pressure $P$ reduces the compressive load carried across grain contacts or an accessible fault. In the usual elementary approximation,At fixed total stress, increasing $P$ translates the entire Mohr circle left by $P$ without changing its radius or the differential stress. The circle therefore approaches the frictional failure envelope even though the applied shear has not increased. Equivalently, in total-stress coordinates the strength on an existing plane is $c+{\mu }_f({\sigma }_n-P)$, reduced by ${\mu }_{f}P$. For a known orientation and constant frictional parameters, the ideal critical pressure isprovided the plane is in contact and sliding failure is reached before tensile opening.

For the diagram's illustrative ${\sigma }_1=80$, ${\sigma }_3=20$, $c=10\mathrm{MPa}$ and ${\mu }_f=0.6$, the dry circle is below failure. The freely oriented tangency threshold is about $P=8.4\mathrm{MPa}$; the $15\mathrm{MPa}$ circle is already beyond that threshold and shows the loading tendency, not a stable post-failure state. If $P$ grows enough that ${\sigma }_3-P\le -T_0$, fluid pressure can instead open a tensile fracture. A porous-continuum effective stress law may use ${\sigma }_i^{\prime }={\sigma }_i-\alpha P$ with a coefficient $\alpha$ different from one. Also, changing fluid pressure can change total stress through constrained deformation; pure horizontal translation assumes those total stresses remain fixed.^[ref]