4
$\begingroup$

Let $\mathbb{R}^\omega$ be the set of all sequences of real numbers with the uniform metric $\tilde{\rho}$ defined as $$ \tilde{\rho}(x,y) \ \colon= \ \sup \left\{ \ \min \left\{ \ \vert x_n - y_n \vert, 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} \ \forall \ x \colon= (x_n)_{n \in \mathbb{N}}, \ y \colon= (y_n)_{n \in \mathbb{N}} \in \mathbb{R}^\omega. $$

Let $a \colon= (a_n)_{n \in \mathbb{N}}$, $b \colon= (b_n)_{n \in \mathbb{N}} \in \mathbb{R}^\omega$ be two fixed sequences with $a_n > 0$ for all $n \in \mathbb{N}$.

Let the map $h \colon \mathbb{R}^\omega \to \mathbb{R}^\omega$ be defined by $$ h\left( (x_n)_{n\in \mathbb{N}} \right) \ \colon= \ \left( a_n x_n + b_n \right)_{n \in \mathbb{N}} \ \forall \ (x_n)_{n \in \mathbb{N}} \in \mathbb{R}^\omega. $$

Then what are the necessary and sufficeint conditions on the $a_n$ and $b_n$ for $h$ to be (i) continuous? (ii) a homeomorphism?

My effort:

If the sequence $(a_n)$ is bounded above by a positive real number $M$, say, then for all $x$, $y \in \mathbb{R}^\omega$ such that $\tilde{\rho}(x,y) < 1$, we have $$ \vert x_n - y_n \vert = \min \{ \vert x_n - y_n \vert , 1\} $$ for each $n \in \mathbb{N}$. And so $$ \tilde{\rho}\left(h(x), h(y)\right) = \sup_{n} \left\vert (a_n x_n + b_n ) - (a_n y_n + b_n) \right\vert = \sup_n \left( \vert a_n \vert \cdot \vert x_n - y_n \vert \right) \leq \sup_n M \vert x_n - y_n \vert = M \sup_n \vert x_n - y_n \vert = M \tilde{\rho}(x,y), $$ from which the continuity of the map follows.

Am I right? And if so, then what is the necessary condition for continuity of $h$?

The map $h$ is bijective.

If in addition the sequence $(a_n)$ is bounded below by a positive real number $m$, say, then using a procedure similar to the one used to obtain the above chain of inequalities, we can arrive at $$ \frac{1}{m} \tilde{\rho}\left( h(x), y(y) \right) \geq \tilde{\rho}(x,y), $$ which shows that the map $h$ is also a homeomorphism.

Am I right? And if so, then what is the necessary condition for the map $h$ to be a homeomorphism?

$\endgroup$

2 Answers 2

1
$\begingroup$

Suppose $(a_n)_n$ is not bounded, then for every $n \in \mathbb{N}$, we find $ m \in \mathbb{N}$ with $a_m \geq n$. Let $x^{(n)} \in \mathbb{R}^\omega$ be defined by $x^{(n)}_{m'} = 0$ for $m' \neq m$ and $x^{(n)}_{m} = a_m^{-1}$. Then $x^{(n)}$ converges with respect to $\tilde\rho$ to $0 \in \mathbb{R}^\omega$, the sequence consisting only of zeroes. But we have $$\tilde{\rho}(h(x^{(n)}),h(0)) \geq |a_m x^{(n)}_m + b_m - b_m| = |a_m x^{(n)}_m| = 1,$$ so $h(x^{(n)})$ does not converge to $h(0)$, thus $h$ is not continuous in this case. By what you have already proved, boundedness of $(a_n)_n$ is a necessary and sufficient criterion for continuity of $h$.

For $h$ to be a bijection you have to have $a_n \neq 0$ for all $n$ which is satisfied by your assumptions. Then the inverse is given by $h^{-1}((x_n)_n) = (a_n^{-1}x_n - a_{n}^{-1}b_n)$. Setting $\tilde{a} = (a_n^{-1})_n$ and $\tilde{b} = (-a_n^{-1}b_n)_n$ we find $$h^{-1}((x_n)_n) = (\tilde{a}_n x_n + \tilde{b}_n)$$ thus we can apply our result concerning the continuity of $h$ to conclude that $h^{-1}$ is continuous iff $\tilde{a_n}$ is bounded, that is, iff $a_n$ is bounded below by some positive number.

$\endgroup$
0
$\begingroup$

Yes you are right.

Let's suppose $b_i=0$. And for if the map is continuous it mis be bounded above, since we're it not, we can find $a_{k_i}$ such that $a_{k_i}>i, i=1, 2, 3,\cdots$, then the sequence $(0,\cdots,1/1,\cdots,1/2,\cdots,1/3,\cdots),(0,\cdots,0,\cdots,1/2,\cdots,1/3,\cdots),(0,\cdots,0,\cdots,0,\cdots,1/3,\cdots),\cdots$ which converges to $(0, 0, 0, \cdots)$ is mapped to a sequence with the distance between every of its member and $(0, 0, 0, \cdots)$ always above $1$.

For the home improvise thing, since they are already bijective, we only need to ensure the inverse is continuous, which means $1/a_i$ is bounded above, that is, $a_i$ is bounded below.

$\endgroup$
1
  • $\begingroup$ can you please add some more detail to your answer? You might even want to see if there's any correction to be made too. $\endgroup$ Apr 20, 2015 at 6:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .