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I am doing the exercises in the book Topology(2nd edition) by Munkres. Here is my question(page 127, question 4(a)):

Let $h:R\to R^\omega$ be a function defined by $h(t)=(t, t/2, t/3, \ldots)$ where $R^\omega$ is in the uniform topology. Is $h$ continuous?

I have been able to determine that $h$ is continuous in the product but not continuous in the box topology. However, I cannot then deduce what will happen in the uniform topology since

"product $\subset$ uniform $\subset$ box"

does not help in this direction.

Please, help. Thank you.

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You're inconsistent with $f$ and $h$. – Asaf Karagila Oct 4 '11 at 10:39
Thanks I have edited it. – math 101 Oct 4 '11 at 11:20
1… – Henno Brandsma Oct 4 '11 at 17:46

A typical basic open nbhd of $h(t)$ in the uniform topology is $$B(h(t),r) = \prod_{k\in\mathbb{Z}^+}\left(\frac{t}{k}-r,\frac{t}{k}+r\right),$$ where $r$ is any positive real number.

Correction: This isn’t quite correct: the sequence $$\left(\frac{t}{k}+\frac{k-1}{k}r\right)_{k\in\mathbb{Z}^+}$$ is in the product of open intervals but not in the open ball of radius $r$ about $h(t)$. The correct definition: $$B(h(t),r) = \bigcup_{0<\epsilon<r}\;\;\prod_{k\in\mathbb{Z}^+}\left(\frac{t}{k}-\epsilon,\frac{t}{k}+\epsilon\right)$$

For $s\in\mathbb{R}$, $h(s) \in B(h(t),r)$ if and only if there is a positive $\epsilon<r$ such that $$\frac{s}{k} \in \left(\frac{t}{k}-\epsilon,\frac{t}{k}+\epsilon\right)$$ for each $k \in \mathbb{Z}^+$. What happens if $|s-t|<r$? (I’ll expand this to a complete solution if you get completely stuck, but you should first try to finish it from here.)

Added: Let’s look at a specific example: the nbhd $B\left(h\left(\frac34\right),\frac14\right)$ of $h\left(\frac34\right)=\left(\frac34,\frac38,\frac3{12}\dots\right)$, the sequence whose $k$-th term is $\frac3{4k}$.

Is $\frac12 \in h^{-1}\left[B\left(h\left(\frac34\right),\frac14\right)\right]$? No: $h\left(\frac12\right)= \left(\frac12,\frac14,\frac16,\dots\right)$, and $\frac12 \notin \left(\frac34-\epsilon,\frac34+\epsilon\right)$ for any $\epsilon<\frac14$. What about $\frac58$? $$h\left(\frac58\right) = \left(\frac58,\frac5{16},\frac5{24},\dots\right),$$ the sequence whose $k$-th term is $\frac5{8k}$. Is it true that there is a positive $\epsilon<\frac14$ such that $$\frac5{8k} \in \left(\frac3{4k}-\epsilon,\frac3{4k}+\epsilon\right)$$ for every positive integer $k$? This is the same as asking whether $$\frac3{4k}-\epsilon<\frac5{8k}<\frac3{4k}+\epsilon$$ for every $k\in\mathbb{Z}^+$. Multiply through by $8k$ and subtract $5$to get $1-8k\epsilon<0<1+8k\epsilon$; as long as $\epsilon>\frac18$, this is true for every positive integer $k$, so $\frac58$ is in $h^{-1}\left[B\left(h\left(\frac34\right),\frac14\right)\right]$.

Now try it in general. Suppose that $|s-t|<r$; is $h(s)\in B(h(t),r)$?

Since $h(s)=(s,s/2,s/3,\dots)$, this is simply asking whether it’s true that there’s a positive $\epsilon<r$ such that $$\frac{s}{k} \in \left(\frac{t}{k}-\epsilon,\frac{t}{k}+\epsilon\right)$$ for all $k\in\mathbb{Z}^+$, i.e., whether $$\frac{t}{k}-\epsilon<\frac{s}{k}<\frac{t}{k}+\epsilon\;.$$ What happens if you pick $\epsilon\in(r-|s-t|,r)$?

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I think $h^{-1}(B(h(t),r))=\{0\}$. Is that correct? If so, then since $\{0\}$ is not open in $\mathbb{R}$, we get that $h$ is not continuous. – math 101 Oct 4 '11 at 11:26
I think $B(h(t),r)$ is not necessarily equal to $\prod_{k\in\mathbb{Z}^+}\left(\frac{t}{k}-r,\frac{t}{k}+r\right)$. Can you please explain? – math 101 Oct 4 '11 at 13:30
@math101: $B(h(t),r)$ is simply a more convenient name that I’m giving to $\prod_{k\in\mathbb{Z}^+}\left(\frac{t}{k}-r,\frac{t}{k}+r\right)$; the two are equal by definition. – Brian M. Scott Oct 4 '11 at 18:35
@math101: My apologies: I misunderstood your second comment. You’re quite correct, if you meant that the product of intervals isn’t the open ball of radius $r$ about $h(t)$; I’ve made the appropriate correction. – Brian M. Scott Oct 4 '11 at 22:50

Since $h$ is linear, and the "uniform" topology is translation invariant, it is sufficient to prove continuity at $0$.

Since $h(0) = 0$, it is sufficient to prove that given any neighborhood $V$ of $0$ in $R^\omega$, $h^{-1}(V)$ will be a neighborhood of $0$. Now, $V$ contains a set of the form $\prod (-\varepsilon,\varepsilon)$ for some $\varepsilon > 0$. So, all one has to check is that $U = h^{-1}\left( \prod (-\varepsilon,\varepsilon) \right)$ contains an interval about $0$. But $U = (-\varepsilon,\varepsilon)$.

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1) $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{uniform})$ is continuous.

Let $\langle x_n\rangle$ be a sequence in $\mathbb{R}$ and let $x\in \mathbb{R}$, and suppose $x_n\to x$ in $\mathbb{R}$ with the usual topology. It suffices to show that $h(x_n)\to h(x)$ in $(\mathbb{R}^\omega,\mbox{uniform})$. Since $h(x_n)$ and $h(x)$ are real valued functions, $h(x_n)$ converges to $h(x)$ uniformly iff $\sup_{k\in\omega}|h(x_n)(k)-h(x)(k)|\to 0$ in $\mathbb{R}$.

$\sup_{k\in\omega}|h(x_n)(k)-h(x)(k)|=\sup_{k\in\omega}|x_n/k-x/k|=|x_n-x|\cdot\sup_{k\in\omega}1/k=|x_n-x|\to 0$ since $x_n\to x$ in $\mathbb{R}$ by assumption.

2) $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{box})$ isn't continuous.

The box topology part is a bit more tedious. Let $x_n$ be a sequence in $\mathbb{R}$ which converges to $x$ in $\mathbb{R}$ with the usual topology. Then $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{box})$ is continuous iff

$\forall\epsilon_1,\epsilon_2,\dots>0,\exists N,\forall n\geq N,\forall k\in\omega,|h(x_n)(k)-h(x)(k)|<\epsilon_k$.

Observe $|h(x_n)(k)-h(x)(k)|=|x_n/k-x/k|=|x_n-x|/k<\epsilon_k$; taking $\epsilon_k=1/k^2$ our expression becomes $\exists N,\forall n\geq N,\forall k\in\omega,|x_n-x|<1/k$, or equivalently $\exists N,\forall n\geq N,|x_n-x|\leq\inf_{k\in\omega}1/k=0$. Choosing a sequence $\langle x_n\rangle$ in $\mathbb{R}$ which isn't eventually constant shows that $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{box})$ isn't continuous.

3) $h:\mathbb{R}\to(\mathbb{R}^\omega,\mbox{product})$ is continuous. [I'm posting this one up since there have been some ugly proofs.]

Again let $x_n\to x$ in $\mathbb{R}$, then it suffices to show that $h(x_n)\to h(x)$ in $(\mathbb{R}^\omega,\mbox{product})$, which holds, iff $h(x_n)(k)\to h(x)(k)$ in $\mathbb{R}$ for all $k\in\omega$. Finally $h(x_n)(k)=x_n/k\to x/k=h(x)(k)$, and we are done.

We can use sequences since $\mathbb{R}$ is first countable.

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