Exercise 20.4

Solution for $(a)$. For the product topology, by Theorem $19.6$, $f,g,h$ are all continuous since each coordinate function is continuous. This is because if an open set in the image of a coordinate function is $(a,b)$, its preimage would still be in the form $(a^{\prime },b^{\prime })\subseteq ℝ$ where $a^{\prime },b^{\prime }$ are determined by the linear equations defining $f,g,h$ above.

Now consider the uniform topology. Note by Theorem $21.1$ we can use the familiar $𝜖$-$\delta$ definition for continuity since our spaces both are metric spaces. We claim $f$ is not continuous. For, suppose it is continuous. Then, given $𝜖>0$ and $x\in ℝ$, there exists $\delta >0$ such that $|x-y|<\delta ⟹|f(x)-f(y)|=\underset{n}{\sup <!--\; FUNCTION\; APPLICATION\; -->}[\min <!--\; FUNCTION\; APPLICATION\; -->(n|x-y|,1)]<𝜖$. But, this is a contradiction since for $n$ large, $\min <!--\; FUNCTION\; APPLICATION\; -->(n|x-y|,1)=1$, and so is always greater than $𝜖$. Now consider $g$. $g$ is continuous since given $𝜖>0$ and $x\in ℝ$, we let $\delta <\min <!--\; FUNCTION\; APPLICATION\; -->(𝜖,1)$ and therefore have $|x-y|<\delta ⟹|f(x)-f(y)|=\underset{n}{\sup <!--\; FUNCTION\; APPLICATION\; -->}[\min <!--\; FUNCTION\; APPLICATION\; -->(|x-y|,1)]=\min <!--\; FUNCTION\; APPLICATION\; -->(|x-y|,1)<\min <!--\; FUNCTION\; APPLICATION\; -->(𝜖,1)\le 𝜖$. $h$ is also continuous since given $𝜖>0$ and $x\in ℝ$, we let $\delta <\min <!--\; FUNCTION\; APPLICATION\; -->(𝜖,1)$ and therefore have $|x-y|<\delta ⟹|f(x)-f(y)|=\underset{n}{\sup <!--\; FUNCTION\; APPLICATION\; -->}[\min <!--\; FUNCTION\; APPLICATION\; -->(|x-y|∕n,1)]\le \min <!--\; FUNCTION\; APPLICATION\; -->(|x-y|,1)<\min <!--\; FUNCTION\; APPLICATION\; -->(𝜖,1)\le 𝜖$.

For the box topology, since the box topology is finer than the uniform topology by Theorem $20.4$, we see that $f$ is not continuous. For, if $V$ open in the uniform topology has preimage that is not open in $ℝ$, $V$ is still open in the box topology and still has the same non-open preimage. Next, by Example $19.2$, we see that $g$ is not continuous. Last, for $h$, we choose

and suppose its preimage $h^{-1}(B)$ is open. This implies $h((-\delta ,\delta ))\subseteq B$, and so applying ${\pi }_n$ gives

for all $n$, a contradiction. □

Solution for $(b)$. We note that since the product topology is Hausdorff by Theorem $19.4$ and both the uniform and box topologies are finer than the product topology by Theorems $19.1$ and $20.4$, if a sequence converges to a point $p$ in one topology, it must converge to the same point in the finer topologies. For, if the sequence converges to $q$ in the finer topology, then it also converges to $q$ in the coarser topology, and by the Hausdorff property $p=q$.

Consider $w_n$. For the product topology, we recall that any basic open set $U=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }\ni 0$ is the product of finitely many open subsets of $ℝ$ with infinitely many copies of $ℝ$. Letting $N$ be the largest $\alpha$ such that $U_{\alpha }⊊ℝ$, we see that $w_n\in U$ for all $n>N$ since the first $N$ components are zero, and the rest are trivially in the remaining copies of $ℝ$ of $U$. Thus, $w_n\to 0$ in the product topology. Now we only have to check if the sequence converges to zero in the other topologies by the above. In the uniform topology, $\overline{\rho }(w_n,0)=1$ for all $n$, and so the sequence does not converge. For, if we choose any ball $U=B(0,r)\subseteq {ℝ}^{\omega }$ for $r<1$, $w_n\notin U$ for all $n$. Finally, since the box topology is finer than the uniform topology by Theorem $20.4$, we see that this same open set $U$ is such that $w_n\notin U$ for all $n$, and so $w_n$ does not converge in the box topology, either.

Consider $x_n$ and $y_n$. We claim they both converge to zero in the uniform topology. For any open set $0\in U\subseteq {ℝ}^{\omega }$ in the uniform topology, we can find $B(0,𝜖)$ such that $B(0,𝜖)\subseteq U$; then, we can find $N$ such that $1∕N<𝜖$. We then see that $x_n,y_n\in B(0,𝜖)\subseteq U$ for all $n\ge N$, and so $x_n,y_n\to 0$ in the uniform topology. Moreover, since the uniform topology is finer than the product topology, we see that this implies $x_n,y_n\to 0$ in the product topology as well. For the box topology, though, we see that neither sequence converges. For, we can construct the set $0\in U={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }(-1∕n,1∕n)$ (where we only consider sets containing zero by the above), which does not contain $x_n,y_n$ for any $n$.

For $z_n$, we see that for any open set $0\in U=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }\subseteq {ℝ}^{\omega }$ in the box topology, for $N$ large enough $1∕n\in U_1,U_2$ for $n\ge N$, and so $z_n\in U$ for all $n\ge N$, since by hypothesis $0\in U$, the third component onwards of $z_n$ are always in their respective $U_{\alpha }$. Thus, $z_n\to 0$ in the box topology; since the box topology is finer than both the uniform and product topologies, we see that this implies $z_n$ converges in the other two topologies as well. □

Proof. The main part of this proof was given after the definition of a metric topology in the text, but we repeat it here for completeness.

By the definition of the metric topology, there is a $\delta >0$ and $y\in X$ such that the basis element $B_d(y,\delta )$ contains $x$ and is contained in $U$. Let $𝜖=\delta -d(x,y)$ so that $d(x,y)=\delta -𝜖$, noting that $𝜖>0$ since $x\in B_d(y,\delta )$ so that $d(x,y)<\delta$. Then, for any $z\in B_d(x,𝜖)$, we have that $d(z,x)<𝜖$ and so

since $d$ is a metric. Hence $z\in B_d(y,\delta )$ so that $B_d(x,𝜖)\subset B_d(y,\delta )\subset U$ as desired since $z$ was arbitrary. □

Proof. For assertion (1) suppose that $f$ is continuous with respect to $Y^{\prime }$ and let $U$ be any open set of $Y$. Since $Y^{\prime }$ is finer than $Y$, it follows that $U$ is also open in $Y^{\prime }$. Then, since $f$ is continuous with respect to $Y^{\prime }$ we have that $f^{-1}(U)$ is open in $X$, which suffices to show that $f$ is continuous with respect to $Y$ since $U$ was an arbitrary open set. Assertion (2) follows immediately from the contrapositive of (1).

Regarding (3), suppose that a sequence $\left(y_1,y_2,\dots \right)$ converges to $y_0$ in $Y^{\prime }$ and let $U$ be any neighborhood of $y_0$ in $Y$. Then $U$ is also open in $Y^{\prime }$ since it is finer than $Y$, hence $U$ is a neighborhood of $y_0$ in $Y^{\prime }$. Thus there is an $N\in {\mathbb{Z}}_{\text{+}}$ such that $x_n\in U$ for all $n\ge N$, since the sequence converges to $y_0$ in $Y^{\prime }$. Since $U$ was an arbitrary neighborhood of $Y$, this shows that the sequence converges to $y_0$ in $Y$. Assertion (4) follows immediately from the contrapositive of (3). Assertion (5) then immediately follows from (4) since, if a sequence does not converge at all in $Y$ then for any point $y_0\in Y$, it does not converge to $y_0$. Then it also does not converge to $y_0$ in $Y^{\prime }$ by (4). Since $y_0$ was arbitrary, this shows that it does not converge at all in $Y^{\prime }$.

For (6), suppose that $Y$ is a Hausdorff space and let $x$ and $y$ be distinct points of $Y^{\prime }$ so that they are of course also points of $Y$. Hence there are neighborhoods $U$ and $V$ of $x$ and $y$, respectively, in $Y$ that are disjoint. Since $Y^{\prime }$ is finer than $Y$, we have that $U$ and $V$ are also open sets of $Y^{\prime }$ and thus are disjoint neighborhoods of $x$ and $y$ in $Y^{\prime }$ as well. This suffices to show that $Y^{\prime }$ is Hausdorff as desired. □

Proof. First, the functions $f$, $g$, and $h$ can all be considered as special cases of the more general function

where each $s_n(t)={\alpha }_{n}t$, and ${\alpha }_n=n$ for $f$, ${\alpha }_n=1$ for $g$, and ${\alpha }_n=1∕n$ for $h$.

Clearly, each $s_n$ is continuous for the three ${\alpha }_n$ by elementary calculus so that $s$ is continuous in the product topology by Theorem 19.6 for all three ${\alpha }_n$. We can show that $s$ is not continuous in the box topology for all three ${\alpha }_n$ with a single example. Consider the set $B={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{n\in {\mathbb{Z}}_{\text{+}}}(-1∕n^2,1∕n^2)$, which is clearly a basis element of the box topology and so is open. Similar to Example 19.2, if $s$ were continuous then there would be an interval $(-\delta ,\delta )$ about the point $0$ such that $s((-\delta ,\delta ))\subset B$, where of course $\delta >0$. This would of course mean that

for all $n\in {\mathbb{Z}}_{\text{+}}$. However, since clearly there is an $n\in {\mathbb{Z}}_{\text{+}}$ large enough that

we have that

and hence for all three functions, we have that ${\alpha }_n\delta >1∕n^2$ so that

This shows that $s$ cannot be continuous with respect to the box topology for all three ${\alpha }_n$.

Next, we show that $f$ is not continuous in the uniform topology. First, suppose that $\overline{\rho }$ is the metric that induces the uniform topology, i.e.

Now consider the basis element and open set $B_{\overline{\rho }}(0,1)$ in the uniform topology. If $f$ were continuous then there would be a $\delta >0$ such that

Clearly there is an $n\in {\mathbb{Z}}_{\text{+}}$ large enough such that $n>1∕\delta$ so that $\mathit{\delta n}>1$. Then consider the point $x\in {ℝ}^{\omega }$ defined by

We then have of course that $-(n+1)\delta <0<\mathit{n\delta }=x_{n+1}<(n+1)\delta$ so that $x_{n+1}\in (-(n+1)\delta ,(n+1)\delta )$. It then follows that $x\in f((-\delta ,\delta )$. However, we also have that $\overline{d}(\mathit{\delta n},0)=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{\left|\mathit{\delta n}-0\right|,1\right\}=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathit{\delta n},1\right\}=1$ since $\mathit{\delta n}>1$. Hence it is not true that $\overline{\rho }(x,0)<1$ so that $x\notin B_{\overline{\rho }}(0,1)$. Thus $f((-\delta ,\delta ))\not\subset B_{\overline{\rho }}(0,1)$ so that $f$ is not continuous in the uniform topology.

Next, we show that $g$ and $h$ are continuous in the uniform topology at the same time, which we show using Theorem 18.1 part (4). Consider any real $u$ and any neighborhood $V$ of $x=g(u)$ (or $x=h(u)$) in the uniform topology. Then by Lemma 1 there is an $𝜖>0$ such that the basis element $B_{\overline{\rho }}(x,𝜖)$ is a subset of $V$. Now consider the basis element and open set $U=B_d(u,𝜖∕2)$, where $d$ denotes the usual metric on $ℝ$. Obviously, $U$ contains $u$ but we also claim that $g(U)\subset V$ (or $h(U)\subset V$), thereby completing the proof.

So consider any $y\in g(U)$ (or $y\in h(U)$) so that there is some $v\in U$ such that $y=g(v)$ (or $y=h(v)$) In the case of $g$ we have that $x=g(u)=(u,u,u,\dots )$, which is to say that $x_n=u$ for all $n\in {\mathbb{Z}}_{\text{+}}$. Similarly $y_n=v$ for all $n\in {\mathbb{Z}}_{\text{+}}$ since $y=g(v)$. Now, since $v\in U=B_d(u,𝜖∕2)$, we have that

for all $n\in {\mathbb{Z}}_{\text{+}}$. From this, it follows that

Likewise in the case of $h$ we have that $x_n=u∕n$ and $y_n=v∕n$ for all $n\in {\mathbb{Z}}_{\text{+}}$ since $x=h(u)$ and $y=h(v)$. We, therefore, have that

for all $n\in {\mathbb{Z}}_{\text{+}}$ since every $n\ge 1$. Hence again

Therefore for both functions we have $\overline{\rho }(y,x)<𝜖$ so that $y\in B_{\overline{\rho }}(x,𝜖)$. This shows that $g(U)\subset B_{\overline{\rho }}(x,𝜖)\subset V$ (or $h(U)\subset B_{\overline{\rho }}(x,𝜖)\subset V$) as desired since $y$ was arbitrary. □

Proof. Now, regarding the $w$ sequence, each element in the sequence is defined as

where

for $n,m\in {\mathbb{Z}}_{\text{+}}$.

First, we show that the $w$ sequence converges to the point $0$ in the product topology. So consider any neighborhood $U$ of $0$ in the product topology so that there is a basis element $B$ containing $0$ where $B\subset U$. Then $B={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{m=1}^{\infty }{B}_m$ where each $B_m$ is open and $B_m$ is all of $ℝ$ for all but finitely many values of $m$. Let $J$ then be a finite subset of ${\mathbb{Z}}_{\text{+}}$ where each $B_m=ℝ$ for $m\notin J$. Of course we also have that $0\in B_m$ for all $m\in {\mathbb{Z}}_{\text{+}}$ since $B$ contains $0$.

Then $J$ has a largest element $N$ since it is a finite set of positive integers. Now consider any $n\ge N+1$ and any $m\in {\mathbb{Z}}_{\text{+}}$. If $m\in J$ then we have that $m\le N<N+1\le n$ since $N$ is the largest element of $J$, and hence $w_{n,m}=0\in B_m$. If $m\notin J$ then of course $B_m=ℝ$ so that of course $w_{n,m}\in ℝ=B_m$ regardless of whether $w_{n,m}=0$ or $w_{n,m}=n$. Hence either way we have $w_{n,m}\in B_m$, which shows that $w_n\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{m=1}^{\infty }{B}_m=B$ since $m$ was arbitrary. Thus also $w_n\in U$ since $B\subset U$. Since $n\ge N+1$ was arbitrary and $U$ was an arbitrary neighborhood of $0$, this shows that the sequence converges to $0$ as desired.

Next, we show that the $w$ sequence does not converge in the uniform topology. It suffices to show that the sequence does not converge to $0$, since if it converged to any other point $x$, then by Lemma 2 part (3) it would also converge to $x$ in the product topology since it is coarser than the uniform topology. However, this would violate the fact that the sequence converges to $0$ in the product topology (just shown above), and so cannot also converge to $x\ne 0$ since the convergence point is unique as noted above.

So consider the neighborhood $B_{\overline{\rho }}(0,1)$ of $0$ in the uniform topology. We claim that no elements of the sequence are in this neighborhood so it clearly cannot converge to $0$. So consider any $n\in {\mathbb{Z}}_{\text{+}}$ so that we clearly have $w_{n,n}=n\ge 1>0$. Therefore $d(w_{n,n},0)=\left|w_{n,n}-0\right|=\left|w_{n,n}\right|=w_{n,n}\ge 1$, from which it follows that it has to be that $\overline{d}(w_{n,n},0)=1$. This of course implies that

Hence it is not true that $\overline{\rho }(w_n,0)<1$ so that $w_n\notin B_{\overline{\rho }}(0,1)$. This shows the desired result since $n$ was arbitrary.

It then follows that the $w$ sequence also does not converge at all in the box topology by Lemma 2 part (5) since it is finer than the uniform topology.

Regarding the $x$ sequence, the definition is that each

where

for $n,m\in {\mathbb{Z}}_{\text{+}}$.

First, we show that this sequence converges to $0$ in the uniform topology, which is of course the unique convergence point. So consider any neighborhood $U$ of $0$ in the uniform topology so that by Lemma 1 there is an $𝜖>0$ where $B_{\overline{\rho }}(0,𝜖)\subset U$. Then there is a positive integer $N$ large enough so that $N>2∕𝜖$ so that, for any $n\ge N$ we have $1∕n\le 1∕N<𝜖∕2$. Next consider any such $n\ge N$ and any $m\in {\mathbb{Z}}_{\text{+}}$. Since $x_{n,m}$ is either $0$ or $1∕n$ we have that $\left|x_{n,m}\right|=x_{n,m}\le 1∕n$ so that

Thus, since $m$ was arbitrary, it follows that

and hence $x_n\in B_{\overline{\rho }}(0,𝜖)$. Thus also $x_n\in U$ since $B_{\overline{\rho }}(0,𝜖)\subset U$. Since $n\ge N$ was arbitrary as was the neighborhood $U$, this shows that the sequence converges to $0$ as desired.

Since it is coarser than the uniform topology, it follows that the $x$ sequence also converges to $0$ in the product topology as well by Lemma 2 part (3).

Next, we show that the $x$ sequence does not converge in the box topology, for which it suffices to show that it does not converge to $0$. Again, this is because, if it were to converge to some other point $y\ne 0$ in the box topology, then it would also converge to $y$ in the uniform topology since it is coarser (Lemma 2 part (3)), but this would violate the fact that it converges to the unique point $0$ by what was just shown. So consider the basis element and open set of the box topology $U={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{U}_n$ where each $U_n=(-1∕n,1∕n)$. Clearly, $U$ contains $0$ so that it is a neighborhood of $0$. We claim that no element of the sequence is in $U$, which of course suffices to show that it cannot converge to $0$. So consider any $n\in {\mathbb{Z}}_{\text{+}}$ and so that $x_{n,n}=1∕n\ge 1∕n$ so that $x_{n,n}\notin (-1∕n,1∕n)=U_n$. From this it follows that $x_n\notin {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{U}_n=U$. Since $n$ was arbitrary this shows no element of the sequence is in $U$ so that the sequence cannot converge to $0$.

Regarding the $y$ sequence, it is defined as

where

for $n,m\in {\mathbb{Z}}_{\text{+}}$. Since $y_{n,m}$ is always either $0$ or $1∕n$, the same argument that shows that the $x$ sequence converges to $0$ in the uniform topology shows that the $y$ sequence does as well. Of course, this also means that it converges to $0$ in the product topology as well since it is coarser. Similarly, the same argument that shows that the $x$ sequence does not converge in the box topology applies to $y$ as well since we have that $y_{n,n}=x_{n,n}=1∕n$ for all $n\in {\mathbb{Z}}_{\text{+}}$.

Now, the $z$ sequence is defined by

where

for $n,m\in {\mathbb{Z}}_{\text{+}}$.

We show that this sequence converges to $0$ in the box topology. So consider any neighborhood $U$ of $0$ in the box topology so that there is a basis element $B={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{m=1}^{\infty }{B}_m$ containing $0$ where $B\subset U$. Of course then each $B_m$ is open in $ℝ$ and $0\in B_m$. Considering the standard topology of $ℝ$ using the metric topology basis, there is then an ${𝜖}_1>0$ such that $B_d(0,{𝜖}_1)\subset B_1$ by Lemma 1 since $B_1$ is open and contains $0$. Likewise, there is an ${𝜖}_2>0$ where $B_d(0,{𝜖}_2)\subset B_2$. So set $𝜖=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{{𝜖}_1,{𝜖}_2\right\}$ so that of course there is a positive integer $N$ large enough that $N>1∕𝜖$. Then, for any $n\ge N$ we have that $n\ge N>1∕𝜖$ so that $1∕n<𝜖\le {𝜖}_1$, and similarly $1∕n<𝜖\le {𝜖}_2$. Now consider any $m\in {\mathbb{Z}}_{\text{+}}$. If $m\le 2$ then of course either $m=1$ or $m=2$ so that, either way, we have

so that $z_{n,m}\in B_d(0,{𝜖}_m)\subset B_m$. If $m>2$ then we clearly have $z_{n,m}=0\in B_m$ as well. Since $m$ was arbitrary, this shows that $z_n\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{m=1}^{\infty }{B}_m=B\subset U$. This shows that the sequence converges to $0$ since $n\ge N$ was arbitrary and $U$ was any neighborhood of $0$.

Of course, this also shows that the $z$ sequence converges to $0$ in the uniform and product topologies as well by Lemma 2 part (3) since they are both coarser than the box topology. □