Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Be $x=(x_{n})$,$y=(y_{n})\in \mathbb{R}^\omega$, be $f\colon [0,1]\subseteq\mathbb{R}\rightarrow \mathbb{R}^\omega$ and $f(t)=(1-t)x_{n}+ty_{n}$.

For $\mathbb{R}^\omega$ with the box topology, show that $f$ is continuous if only if $\exists N\in \mathbb{N}$ that $\forall_{n \geq N} x_{n}=y_{n}$

$\Leftarrow$ no problem,

$\Rightarrow$ I have problems, i think in use continuous properties, like $f(\overline{A})\subseteq\overline{f(A)}$ but dont result.

Any help is appreciated.

$$\mathbb{R}^\omega=\prod_{n=1}^{\infty}{\mathbb{R}}$$

$\mathbb{R}$ usual topology

share|improve this question
    
What is your distance associated with the box topology? –  S4M Jun 19 '12 at 22:33
    
Distance? Box topology is a kind of product topology different from Tychonoff (all products of open sets are open). –  tomasz Jun 19 '12 at 22:35

2 Answers 2

Suppose by contradiction that there are infinitely many $n$ such that $x_n\neq y_n$.

Then we can assume without loss of generality that for all $n$, $x_n<y_n$. For each $n$ take $\varepsilon_n>0$ such that if $x_n(1-t)+y_nt>y_n-\varepsilon_n$, then $t>1-1/n$.

Then the preimage of the open box $\prod (y_n-\varepsilon_n,y_n+\varepsilon_n)$ is $\lbrace 1\rbrace$, which is not open, so it's a contradiction and we're done.

==edit==

Specifically, you could put $\varepsilon_n=(y_n-x_n)/n$. Then we have $$x_n(1-t)+y_nt>y_n-\varepsilon_n=y_n(1-1/n)+x_n/n$$ equivalent to $$x_n(1-t-1/n)+y_n(t-1+1/n)=(t-(1-1/n))(y_n-x_n)>0$$ by $y_n>x_n$ equivalent to $t>1-1/n$.

Anything smaller works as well, of course (except there will be implication and not equivalence, but that's what we need).

share|improve this answer
    
i dont understand u proof, your $\varepsilon_{n}$ how is?. and after the inequalities are strange, you can be more clear please, sorry for my english –  kEoz Jun 19 '12 at 23:39
    
@kEoz I explained in more detail what $\varepsilon_n$ you could choose. –  tomasz Jun 20 '12 at 0:51

Here's some intuition for the problem:

First, think about working coordinate-wise, one $n$ at a time. What does it mean for $(1-t)x_n + ty_n$ to be continuous (at $t_0$)?

It means for all $\epsilon > 0$, there must be a $\delta >0$ such that if $0<|t-t_0|<\delta$, then $|f(t_0) - f(t)|<\epsilon$.

In fact, we can choose $\delta = \epsilon/|x_n - y_n|$ when $x_n\neq y_n$. To see this, just compute: \begin{align*} |f(t_0) - f(t)| &= |(1-t_0)x_n + t_0y_n - (1-t)x_n - ty_n| \\\ &= |(t-t_0)(x_n - y_n)| \\\ &= |t_0-t||x_n - y_n|\\\ &< \frac{\epsilon}{|x_n-y_n|}|x_n-y_n|\\\ &= \epsilon \end{align*}

So we can choose this $\delta$, but we learn something more: this is the best possible $\delta$. For if $|t-t_0| = \delta$, then the same calculation shows that $|f(t) - f(t_0)| = \epsilon \not<\epsilon$.

What happens if $x_n = y_n$? In this case, $(1-t)x_n + y_n = x_n$, so it's constant. This means that given any $\epsilon$, we may choose any $\delta$ we want.

Now, in the box topology, we get to pick a different $\epsilon_n$ on each factor, but the $\delta$ that we pick must work for all $\epsilon_n$ simultaneously. This leads us to the idea of the solution: If there are only finitely many $x_n\neq y_n$, then no matter what $\epsilon_n$s we choose, we can choose the $\delta_n$s as above, and let $\delta = \min\{\delta_n\}$.

On the other hand, if we have $x_n\neq y_n$ infinitely often, then, given $\epsilon_n$s the same approach as in the previous paragraph leads us to try $\delta = \inf\{\delta_n\}$. In fact, if anything works, it must be this $\delta$, owing to the fact that each $\delta_n$ above is the best possible. Note that we're taking an $\inf$ instead of a $\max$ because there are infinitely many $\delta_n$s. Now, if this $\inf$ is $0$, then the function $f$ fails to be continuous.

So, in order to force $\delta = 0$, we need the $\delta_n$s to decrease to $0$, at least when considering the $n$ where $x_n\neq y_n$. To do this, pick $\epsilon_n = \frac{1}{n} |x_n-y_n|$ (if $x_n = y_n$, just pick $\epsilon = 1$).

Then our above formula for $\delta_n$ gives $\delta_n = \frac{\epsilon}{|x_n-y_n|} = \frac{1}{n}$. Then, $\delta = \inf\{\delta_n\} = 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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