- #1
jostpuur
- 2,116
- 19
Let [itex]a\in\mathbb{R}[/itex], [itex]a>0[/itex] be fixed. We define a mapping
[tex]
\mathbb{Q}\to\mathbb{R},\quad q\mapsto a^q
[/tex]
by setting [itex]a^q=\sqrt[m]{a^n}[/itex], where [itex]q=\frac{n}{m}[/itex]. How do you prove that the mapping is locally uniformly continuous? Considering that we already know what [itex]q\mapsto a^q[/itex] looks like, we can define the local uniform continuity by stating that the restriction to [itex][-R,R]\cap\mathbb{Q}[/itex] is uniformly continuous for all [itex]R>0[/itex]. The continuity is considered with respect to the Euclidian metric, which [itex]\mathbb{Q}[/itex] inherits from [itex]\mathbb{R}[/itex].
The use of a mapping
[tex]
\mathbb{R}\to\mathbb{R},\quad x\mapsto a^x
[/tex]
and its derivative is not allowed, because the claim is elementary, and may be needed in the proofs of the most basic results concerning the [itex]a^x[/itex].
edit: Actually I don't know how to prove that [itex]q\mapsto a^q[/itex] is merely continuous in the ordinary way either, so I wouldn't mind some information on that too.
[tex]
\mathbb{Q}\to\mathbb{R},\quad q\mapsto a^q
[/tex]
by setting [itex]a^q=\sqrt[m]{a^n}[/itex], where [itex]q=\frac{n}{m}[/itex]. How do you prove that the mapping is locally uniformly continuous? Considering that we already know what [itex]q\mapsto a^q[/itex] looks like, we can define the local uniform continuity by stating that the restriction to [itex][-R,R]\cap\mathbb{Q}[/itex] is uniformly continuous for all [itex]R>0[/itex]. The continuity is considered with respect to the Euclidian metric, which [itex]\mathbb{Q}[/itex] inherits from [itex]\mathbb{R}[/itex].
The use of a mapping
[tex]
\mathbb{R}\to\mathbb{R},\quad x\mapsto a^x
[/tex]
and its derivative is not allowed, because the claim is elementary, and may be needed in the proofs of the most basic results concerning the [itex]a^x[/itex].
edit: Actually I don't know how to prove that [itex]q\mapsto a^q[/itex] is merely continuous in the ordinary way either, so I wouldn't mind some information on that too.
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