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We consider the cartesian product $$X=\prod_{s\in [2,3]}[0,1]=\{(t_s)_{s\in [2,3] }:t_s\in [0,1]\}.$$

I will just write $(t_s)$ instead of $(t_s)_{s\in [2,3]}$.

Let $\tau_b$ be the box topology where the open basic subsets are $\prod_{s\in [2,3]}U_s$ where each $U_s\subseteq [0,1]$ is open, and $\tau_\phi$ the topology induced by the uniform metric $$\phi((t_s),(y_s))=\sup\{|t_s-y_s|:s\in [2,3]\}.$$

I want to prove that $\tau_{\phi}\subseteq\tau_b$. I think I almost have it, but still stuck in some part.

Let $B((t_s),\epsilon)$ be an open ball in $(X,\tau_{\phi})$ and $(y_s)\in B((t_s),\epsilon)$. In order to show that $(y_s)$ is an interior point of $B((t_s),\epsilon)$ in the space $(X,\tau_b)$, I must show there is some open basic subset in the box topology between them.

First I tried with $\delta = \epsilon -\phi ((t_s),(y_s))$, and wanted to show if $$\prod_{s\in [2,3]}(y_s-\delta,y_s+\delta)\subseteq B((t_s),\epsilon).$$

But it won't work because if $(z_s)$ is in that product, then for each $s\in [2,3]$, we have $|z_s-y_s|<\delta $ and then $$|z_s-t_s|\le|z_s-y_s|+\phi((t_s),(y_s))<\epsilon,$$

and this only takes us to $\phi((z_s),(t_s))\le\epsilon$, so we cannot say that $(z_s)$ is in the open ball $B((t_s),\epsilon)$.

Any hint? Thank you.

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Hint: Try using a smaller $\delta$. If you replace $\delta$ with some $\delta'<\delta$, then at the end you'll get $|z_s-t_s|<\epsilon-(\delta-\delta')$, which is a bound on $|z_s-t_s|$ independent of $s$ which is $<\epsilon$.

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  • $\begingroup$ Tried with $\delta /2$, but I only had $|z_s-y_s|+\phi /2 <\epsilon /2$. I don't know how to conclude $|z_s-t_s|<\epsilon /2$. It need to be a smaller $\delta$? $\endgroup$ – JonSK Jan 6 '16 at 3:13
  • $\begingroup$ Just use $\delta/2$ and use the same inequality $|z_s-t_s|\leq |z_s-y_s|+\phi$ you used above. You then get $|z_s-y_s|+\phi<\delta/2+\phi$. Can you show this is $<\epsilon$? $\endgroup$ – Eric Wofsey Jan 6 '16 at 3:18
  • $\begingroup$ Oh I see, and $\delta/2 + \phi$ is independent of $s$. I understand now. Thank you. $\endgroup$ – JonSK Jan 6 '16 at 3:32
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Every non-empty open ball of $\tau_{\phi}$ belongs to $\tau_b$.( Since the set of non-empty open $\tau_{\phi}$-balls is a base for $\tau_{\phi}$ this gives $\tau_{\phi}\subset \tau_b)$. .....PROOF: For $y=(y_i)\in X$, for $r>0,$ for $ x=(x_i)\in X$ we have $x\in B_{\phi}(y,r)$ $\iff \sup_i |x_i-y_i|<r $ $ \iff \exists n\in N \;\forall i\;(|x_i-y_i|<r(1-2^{-n}))\iff$ $\iff x\in \cup_{n\in N}V_n\;$ where $V_n= \prod_i W_{i,n} \;$ where $$W_{i,n}= [0,1]\cap (y_i-r(1-2^{-n}),y_i+r(1+2^{-n})).$$ Each $W_{i,n}$ is open in $[0,1]$ so each $V_n\in \tau_b$...So $B_{\phi}(y,r)=\cup_{n\in N}V_n\in \tau_b$.

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