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Let $\mathbb{R}^\omega$ denote the set of all sequences of real numbers, and let $(a_1, a_2, a_3, \ldots ), (b_1, b_2, b_3, \ldots) \in \mathbb{R}^\omega$ be fixed with $a_i > 0$ for all $i= 1, 2, 3, \ldots$. Let the map $h \colon \mathbb{R}^\omega \to \mathbb{R}^\omega$ be defined as follows: $$h(x_1, x_2, x_3, \ldots) \colon= (a_1x_1 + b_1, a_2 x_2 + b_2, a_3 x_3 + b_3, \ldots).$$

Then $h$ is a homeomorphism if $\mathbb{R}^\omega$ be given the product topology.

What is the situation in case $\mathbb{R}^\omega$ is given the box topology?

My effort:

The map $h$ is also a homeomorphism if $\mathbb{R}^\omega$ be given the box topology.

Let $B \colon= \Pi_{i \in \mathbb{N}} (c_i, d_i) $ be any basis element for the box topology, where $$(c_i, d_i) \colon= \{ \ x_i \in \mathbb{R} \ c_i < x_i < d_i \ \}$$ is an open interval on the real line, for each $i$.

Then since $a_i > 0$ for each $i$, we have $$h(B) = \Pi_{i \in \mathbb{N}} (a_i c_i + b_i , a_i d_i + b_i ) $$ which is open in the box topology on $\mathbb{R}^\omega$ (in fact an element of a basis). Thus $h$ is an open map.

And, $$h^{-1}(B) = \Pi_{i \in \mathbb{N}} \left( \frac{c_i - b_i}{a_i}, \frac{d_i - b_i}{a_i} \right), $$ which is again open. Thus $h$ is continuous.

Am I right?

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    $\begingroup$ Are you sure that is the title you want for your question?? Because it already is a serious candidate to "Weirdest title ever in ME" . $\endgroup$ – Timbuc Mar 31 '15 at 10:51
  • $\begingroup$ Oh sorry! But that was not the title I'd intended to give this question. I just can't figure out how my entry got replaced by that funny phrase. $\endgroup$ – Saaqib Mahmood Mar 31 '15 at 11:25
  • $\begingroup$ Everything you wrote is correct and well formulated. Good job. $\endgroup$ – Crostul Mar 31 '15 at 13:56
  • $\begingroup$ @Crostul Thanks. $\endgroup$ – Saaqib Mahmood Mar 31 '15 at 14:21

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