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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.


$\mathbb{R}$ usual topology

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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

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.


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).

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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, $\delta = \inf\{\delta_n\} = 0$.

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