A space is path-connected if for every pair $x_0,x_1$ it contains a continuous path $\gamma :[0,1]\to X$ with $\gamma (0)=x_0$ and $\gamma (1)=x_1$. For any $y_0,y_1\in f(X)$ choose preimages $x_0,x_1$. A path between those preimages maps to $f\circ \gamma$, a continuous path from $y_0$ to $y_1$ staying in $f(X)$. Thus a continuous image of a path-connected space is path-connected. Surjectivity onto the ambient codomain $Y$ is not needed; the conclusion concerns $f(X)$.

A topological space is connected if it cannot be expressed as a union of two disjoint nonempty sets open in that space. Suppose $f(X)$ had such a separation $U\cup V$. Here $U,V$ are open in the subspace topology of $f(X)$. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$. They are disjoint, nonempty because $U,V$ contain image points, and cover $X$. This contradicts connectedness. Hence a continuous image of a connected space is connected.

By part (ii) and completeness of each factor, choose $x_i\in X_i$ with $x_i^{(n)}\to x_i$. Let $x=(x_i)$. Given $\epsilon >0$, choose $N$ such that ${\sum }_{i>N}{2}^{-i}=2^{-N}<\epsilon /2$. Coordinate convergence gives a single $n_0$ for which $d_i(x_i^{(n)},x_i)<\epsilon /2$ for every $i\le N$ and $n\ge n_0$. Since $\varphi (t)\le t$,Thus $x^{(n)}\to x$ in the product metric, proving the product is complete. The finite head uses coordinate convergence; the summable tail is controlled uniformly, regardless of how large the remaining original coordinate distances are. This is completeness of a countable product metric.

Let $(x^{(n)})$ be Cauchy for $d$, and fix a coordinate $i$. Every nonnegative summand is bounded by the whole sum:Given $\epsilon >0$, choose $N$ such that the right-hand side is less than $2^{-i}\epsilon /(1+\epsilon )$ for all $m,n\ge N$. Then strict monotonicity of $\varphi$ implies $d_i(x_i^{(m)},x_i^{(n)})<\epsilon$. Hence every coordinate sequence is Cauchy in its original metric $d_i$, with the index bound allowed to depend on the fixed coordinate.

Put $\varphi (t)=t/(1+t)$. Every summand is nonnegative and at most $2^{-i}$, so the series defining $d$ converges and $d\le 1$. Symmetry follows from each $d_i$. If $d(x,y)=0$, every nonnegative summand is zero, hence every $d_i(x_i,y_i)=0$ and $x_i=y_i$. The converse is immediate.

For the triangle inequality, first observe that for $a,b\ge 0$,Monotonicity of $\varphi$ and the triangle inequality for $d_i$ therefore giveMultiply by $2^{-i}$ and sum the convergent nonnegative series. This proves the required triangle inequality for $d$, so $d$ is a metric. The construction combines a bounded metric transform with a countable product metric.

A complete metric space is one in which every Cauchy sequence converges to a point of that space. For an incomplete example, take $(0,1)$ with the usual Euclidean metric and $x_n=1/(n+1)$. It is Cauchy because it converges to zero in $\mathbb{R}$. It cannot converge inside $(0,1)$: any such convergence would also hold in $\mathbb{R}$, contradicting uniqueness of the limit, since zero is excluded. Thus the ambient metric and the underlying set both matter to completeness.

The sequence converges to $x\in X$ ifEquivalently $d(x_n,x)\to 0$. The limit must belong to the specified metric space, not just to a larger space containing it.

A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence ifThe index $N$ may depend on $\epsilon$, but must work for every pair of sufficiently late terms. No candidate limit is part of this definition.

Put $D(z)={\sinh }^{2}z+{\cosh }^{2}a$ and $S=\sinh (2a)>0$. Traverse the rectangle $-R\to R\to R+\pi i\to -R+\pi i\to -R$ with positive orientation of a contour, taking $R>a$. Part (i) shows that the only enclosed poles of $f(z)=z/D(z)$ areThey are simple poles becauseThe residue formula for a quotient at a simple denominator zero givesTheir imaginary terms cancel and their residues sum to $-2a/S$. Thus Cauchy's residue theorem gives

The choice of numerator $z$ makes the horizontal sides useful. The hyperbolic sine satisfies $\sinh (z+\pi i)=-\sinh z$, so $D(x+\pi i)=D(x)$. Consequently the lower and upper contour integrals, with their actual orientations, add toIt remains to justify discarding both vertical sides. On either side $z=\pm R+iy$, $0\le y\le \pi$, the reverse triangle inequality and part (b)(ii) implyThis lower bound is positive for sufficiently large $R$. Also $|z|\le R+\pi$ and each vertical side has length $\pi$. The estimation lemma bounds the absolute value of the sum $V_R$ of the vertical contributions byThe real improper integral converges absolutely, since its integrand is continuous on the real axis and decays like $4e^{-2|x|}$ at infinity. Combining all four sides and taking $R\to \infty$ givesThereforeAn independent real-variable check uses the hyperbolic tangent substitution $u=\tanh x$: the integral becomes ${\int }_{-1}^{1}du/({\cosh }^{2}a-u^2{\sinh }^{2}a)$. Its antiderivative uses the inverse hyperbolic tangent, giving $2\mathrm{arctanh}(\tanh a)/(\sinh a\cosh a)=4a/\sinh (2a)$. As $a\downarrow 0$, this tends to $2$, agreeing with ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{2}xdx=2$ and checking the factor of two.

The equation requires $\sinh z=\pm i\cosh a$. Write $z=x+iy$, with real $x,y$. Comparing real parts and imaginary parts using part (b)(i) givesIf $x=0$, the second equality would require $|\sin y|=\cosh a>1$, impossible because $a>0$. Thus $\sinh x\ne 0$, forcing $\cos y=0$ andNow $|\sin y|=1$, so $\cosh x=\cosh a$. The hyperbolic cosine is even and is a strictly increasing function on $[0,\infty )$, hence $x=\pm a$. Conversely, substituting these values gives $\sinh (x+iy)=i(-1{)}^k\cosh a$, so every value found does satisfy the original equation. Therefore the complete set is

The two terms in the preceding formula are the real part and imaginary part. Their squared sum is the squared complex modulus:Here ${\cosh }^{2}x-{\sinh }^{2}x=1$ follows directly from the exponential function definitions. The hyperbolic sine modulus identity therefore yieldsEquality holds exactly when $\sin y=0$, that is, when $y$ is an integer multiple of $\pi$. The lower bound is uniform in $y$, which will control whole vertical sides of the contour rather than just their endpoints.

Start with the exponential function definition of the hyperbolic sine and use Euler's formula:Identifying the hyperbolic sine and hyperbolic cosine in the final line gives

Use the generalized binomial theorem with exponent $-1/2$, choosing the analytic branch of a square root that equals $1$ at $z=0$. For $|z|<1$,The signs cancel in each binomial coefficient. For $n\ge 1$,The same expression equals $1$ for $n=0$, using the empty-product convention. Consequently the power series isIts coefficients are the normalized central binomial coefficients. As a convergence check, the ratio of successive absolute terms tends to $|z|$, so the ratio test gives absolute convergence inside the unit disc. Applying the Cauchy coefficient formula to the holomorphic function $(1-z{)}^{-1/2}$ now immediately yieldsThis value is independent of the chosen radius because the contour integral extracts the same coefficient on every such circle.

Choose $0<\rho <r$ and traverse the circle $|z|=\rho$ counterclockwise, so it has positive orientation of a contour. The Cauchy coefficient formula gives, for every nonnegative integer $n$,The smaller radius matters: the holomorphic function is only known on $|z|<r$, so its values on $|z|=r$ need not exist. To see directly why this contour integral selects $c_n$, the power series converges uniformly on $|z|=\rho$, allowing uniform convergence and contour integration to justify integrating term by term. The contour integral of $z^{k-n-1}$ is zero for $k\ne n$ and $2\pi i$ for $k=n$. Thus all other terms disappear.

Write $C^{\prime }=C\setminus \{(1,1)\}$ and $V=\{1\}\times [0,1)$. Let $L$ and $R$ be the portions of $C^{\prime }$ with first coordinate respectively less than and greater than one. Each is path-connected: move along a rational tooth to its corresponding open half of the top line, then horizontally there. By the density of the rational numbers, every point of $V$ is a limit point of both $L$ and $R$.

Use the elementary fact that adjoining any collection of limit points to a connected set preserves connectedness. Indeed, in a proposed separation, the connected subset lies on one side; relative openness would prevent any point on the other side from being a limit point of it. Therefore $L\cup V$ and $R\cup V$ are connected. They intersect in $V$, so their union $C^{\prime }$ is connected.

There is, however, no path from $(1,0)$ to $(0,1)$. If $\gamma (t)=(x(t),y(t))$ were one, let $t_0$ be the first time at which $y(t_0)=1$. Such a first time exists by continuity and compactness of $[0,1]$, and $t_0>0$. For $0\le t<t_0$, membership in the comb forces $x(t)\in \mathbb{Q}$. The continuous image of a connected space is connected, and a connected subset of $\mathbb{Q}$ is a singleton: an irrational number between any two distinct rational values would separate it. Hence $x(t)=1$ for $t<t_0$. By continuity, $\gamma (t_0)=(1,1)$, which was deleted.

Let $D$ be this union of rational-coordinate horizontal and vertical lines. A point $(x,y)\in D$ has either rational $x$, in which case it moves vertically to $(x,0)$ then horizontally to $(0,0)$, or rational $y$, in which case it moves horizontally to $(0,y)$ then vertically to $(0,0)$. Every segment remains on a line with a rational fixed coordinate.

Thus $D$ is path-connected and connected. It is not necessary for both coordinates to be rational.

Each vertical line segment over a rational coordinate meets the horizontal line at height one. A point on such a segment can move vertically to the top, then horizontally along that line; points already at height one use only the horizontal motion. Concatenating these motions joins any two points.

Thus the rational comb with a deleted attachment point construction, before deletion, is path-connected and hence connected. The infinitely many vertical pieces cause no difficulty because each individual path uses only finitely many segments.

Only the right-hand ball is closed, but the union is still path-connected and connected. Its right-hand closed ball contains zero, and every point $z$ of the left-hand open ball can be joined to zero by $\gamma (t)=(1-t)z$.

To check the endpoint carefully, write $a=(-1,0)$. For $0\le t<1$, convexity and the triangle inequality giveAt $t=1$, the path reaches zero in the other, closed ball. Points in that ball also join zero by line segments, completing the construction.

Each closed ball is a convex set and contains the origin. Join any point to zero by a line segment inside its ball, then join zero to the other point. Concatenating the two segments gives a path in the union. It is therefore path-connected and connected.

The two open balls are disjoint: a common point would have distances to their centres summing to less than two, contradicting the triangle inequality. Both are nonempty and relatively open, and they cover the union. They therefore give a separation, so the union is neither connected nor path-connected.