Exercise 25.2

Solution for $(a)$. By Exercise 24.8(a), ${ℝ}^{\omega }$ is path connected, for Theorem 19.6 is not limited to finite product topologies. Thus, ${ℝ}^{\omega }$ is the only path component, and so ${ℝ}^{\omega }$ is the only component as well since path connected $⟹$ connected. □

Proof of $(b)$. We first define $\phi :x\mapsto x-y$. We recall that since $\overline{\rho }(\phi (x),\phi (z))=\overline{\rho }(x,z)$, by Exercise 21.2 $\phi$ is an isometric imbedding that is moreover surjective (the preimage of any $z$ is $z+y$), $\phi$ is a homeomorphism. Thus, $x-y$ is in the same component as $0$ if and only if $x$ is in the same component as $y$, for $\phi ,{\phi }^{-1}$ do not modify the topology of ${ℝ}^{\omega }$.

It therefore suffices to check the case $y=0$. Suppose $x$ is bounded; then, we define $f:[0,1]\to {ℝ}^{\omega }$ where $f(t)=(x_{1}t,x_{2}t,\dots )$. This is continuous since given $𝜖>0$, $B(f(t),𝜖)\supseteq f(B(t,𝜖∕\sup <!--\; FUNCTION\; APPLICATION\; -->\{|x_n|\}))$, where $\sup <!--\; FUNCTION\; APPLICATION\; -->\{|x_n|\}<\infty$ by boundedness of $x$. Thus, $f$ connects $0$ and $x$, i.e., they are in the same path component, and therefore the same component by Theorem 25.5.

Conversely, recall by Exercise 23.8 that we have the separation ${ℝ}^{\omega }=A\cup B$, where $A$ is the set of bounded sequences and $B$ is the set of unbounded sequences of reals. If $x$ is unbounded it is in $B$ and so is not in the same component as $0$. □

Proof of $(c)$. $x$ “eventually zero” here means that $x_i=0$ for all $i\ge N$ for some $N$. Note by the same argument as in $(b)$, it suffices to consider the case $y=0$.

Suppose first that $x$ is not eventually zero. Define the function $f=(f_n)$, where $f_n(a)=\mathit{na}∕\left|x_n\right|$ if $x_n\ne 0$, and $a$ otherwise. $f$ is continuous since each $f_n$ is continuous since it is linear, and so if $f_n^{-1}(U_n)=V_n$, we have $f^{-1}(\prod <!--\; FUNCTION\; APPLICATION\; -->U_n)=\prod V_n$. Note that this is a bijection since each component has an inverse $f_n^{-1}(a)=\left|x_n\right|a∕n$ if $x_n\ne 0$, and $a$ otherwise, and moreover since the inverse is continuous since each component is linear, we have a homeomorphism $f:{ℝ}^{\omega }\to {ℝ}^{\omega }$. Since there are infinitely many $n$ such that $x_n\ne 0$, and so infinitely many $n$ such that $f_n(x_n)=n$, we have that $f(x)$ is unbounded, and thus, by the separation of ${ℝ}^{\omega }$ in the box topology in Example 23.6, we have that $f(x)$ and $0$ are in different components. Since $f$ is a homeomorphism, this implies $x$ and $0$ are in different components as well.

Conversely, suppose $x$ is eventually zero. Then, $x_n=0$ for all $n\ge N$ for some $N$, and so $x\in {ℝ}^N\times \{0\}\times \{0\}\times \cdots \subseteq {ℝ}^{\omega }$; this subspace is homeomorphic to ${ℝ}^N$. Since ${ℝ}^N$ is connected by Theorem $23.6$, we see that $x$ and $0$ are in the same component. □