a) Consider a continuous function $f:\mathbb{R}^\mathbb{N}\rightarrow \mathbb{R}$ defined on $\mathbb{R}^\mathbb{N}$ with the Tychonoff topology, such that $\left (\forall q\in \mathbb{Q}^\mathbb{N}\right) f(q)=\cot^{-1} {q_{2016}}$. Prove that $\left(\forall x \in \mathbb{R}^\mathbb{N} \right) f(x)=\cot^{-1} x_{2016}$.

b) Does the same hold if the topology on $\mathbb{R}^\mathbb{N}$ is the box topology?

I started by taking an arbitrary sequence $x=\{x_n\}_{n\in\mathbb N}$ and a sequence of sequences $\{q_n\}_{n\in\mathbb N}=\left \{\{q_n^k\right \}_{k\in\mathbb N}\}_{n\in\mathbb N}$ such that $x=\lim\limits_{n\to\infty}q_n$. This means that for every basic neighbourhood of $x$, all the sequences $q_n$ will be in that neighbourhood of $x$ except finitely many.

We now take an artbitrary open interval $U\ni x$ and the neighbourhood $\prod_{i=1}^\infty X_i$ of $x$, where $(\forall i\in\mathbb N\setminus\{2016\}) X_i=\mathbb R$, and $X_{2016}=U$. Since this is an open neighbourhood of $x$, all but finitely many $q_n$ are in this neighbourhood. This means that for any basic neighbourhood $U$ of $x_{2016}$, all but finitely many $q_n^{2016}$ are in $U$, i.e., $x_{2016}=\lim\limits_{n\to\infty} q_n^{2016}$.

Now, since $f$ is continuous, $$f(x)=f\left(\lim\limits_{n\to\infty}q_n\right)=\lim\limits_{n\to\infty}f(q_n)=\lim\limits_{n\to\infty}\cot^{-1}q_n^{2016}=\cot^{-1}\left( \lim\limits_{n\to\infty} q_n^{2016} \right)=\cot^{-1}x_{2016}.$$ Q.E.D.

Is this proof correct? Also, regarding the second part, I fail to see where the proof breaks up for the box topology, despite my initial guess that the statement would not hold in that case. Is it correct, or is there a counterexample for it?

  • $\begingroup$ I have deleted my A because I'm confused by the appearance of $q_{2016}$ and $q_n^{2016}$. Is $q_{2016}$ the $2016$-th co-ordinate of $q$? And what then is $q_n^{2016}$? $\endgroup$ – DanielWainfleet Nov 23 '16 at 2:04
  • $\begingroup$ @user254665 The question and the proof use different notations. $q_{2016}$ is the 2016th coordinate of $q$, and $q_n^{2016}$ is the 2016th coordinate of $q_n$. $\endgroup$ – ryagami Nov 23 '16 at 2:14
  • $\begingroup$ Thank you.I have re-written and undeleted my answer. $\endgroup$ – DanielWainfleet Nov 23 '16 at 2:48

Theorem: Let $Y$ be Hausdorff . Let $f:X\to Y$ and $g:X\to Y$ be continuous. Then $\{a\in X: f(a)=g(a)\}$ is closed in $X.$ (Not hard, but I think it's easier to show directly that $\{a\in X: f(a)\ne g(a)\}$ is open in $A.$)

Corollary: If $Y$ is Hausdorff and $f:X\to Y,\; g:X\to Y$ are continuous, and $f,g$ agree on a dense subset of $X$, then $f=g.$

$\mathbb Q^{\mathbb N}$ is dense in $\mathbb R^{\mathbb N}$ in the Tychonoff, uniform, and box topologies. The projection $p_{2016}$ of $\mathbb R^{\mathbb N}$ to its $2016$-th co-ordinate is continuous in each of these topologies. And $cot^{-1}$ is a continuous real function. So $g(x)=\cot^{-1}p_{2016}(x)$ is continuous. And $\mathbb R$ is Hausdorff. So if $f:\mathbb R^{\mathbb N}$ is continuous with respect to any of these topologies, and $f(x)=g(x)$ for all $x\in \mathbb Q^{\mathbb N}$ then $f=g.$


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.