3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\epsilon\in(r-|s-t|,r)$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – math 101 Oct 4 '11 at 11:26
  • $\begingroup$ 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? $\endgroup$ – math 101 Oct 4 '11 at 13:30
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Oct 4 '11 at 18:35
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Oct 4 '11 at 22:50
0
$\begingroup$

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 $U = (-\varepsilon,\varepsilon)$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.