I know that $[0,M]\subset R_+ $ is connected, separable. Now, let us consider the infinite dimensional space $[0,M]^\infty $. I want to see whether this space in connected and separable. I think the first step is to think of the relevant distance function. For two elements $x,y\in[0,M]^\infty $, let $d(x,y)=\sup_tx_ty_t$. Is $[0,M]^\infty $ connected and separable?
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Your distance is not well defined. $d((0,0,...), (1,2,3...)) = \infty$. – copper.hat Jul 09 '14 at 03:34

I am sorry as I do not understand. The second element (1,2,3...) is bounded above in each component by M. Hence, what you said should not be a problem. – Juanito Jul 09 '14 at 03:38

@copper.hat: Since every coordinate is bounded by $M$, this metric is bounded. – Eric Towers Jul 09 '14 at 03:38

1Separable is easy: sequences of bounded rationals are dense in this space. – Eric Towers Jul 09 '14 at 03:39

Sorry, wasn't paying attention. – copper.hat Jul 09 '14 at 03:45

The space is convex, hence the line joining any two points lies in the set. – copper.hat Jul 09 '14 at 03:47

After thinking a bit longer, connected is easy: [0,M] is connected. The product of connected spaces is connected. – Eric Towers Jul 09 '14 at 03:47

@copper.hat: Connected $\neq$ pathconnected $\neq$ arcconnected. See http://en.wikipedia.org/wiki/Connected_space – Eric Towers Jul 09 '14 at 03:48

@EricTowers: A little care is needed with infinite products. – copper.hat Jul 09 '14 at 03:48

In fact the **arbitrary** product of connected spaces is connected. http://math.stackexchange.com/questions/257981/productofconnectedspacesproof – Eric Towers Jul 09 '14 at 03:49

@EricTowers: (1) Path connected implies connected (and convex implies path connected). (2) I think the proof given there is for the product topology which is not compatible with the given metric. – copper.hat Jul 09 '14 at 03:51

@copper.hat: This metric is equivalent to the uniform metric, which produces a larger (finer) topology than the product topology. – Eric Towers Jul 09 '14 at 03:54

@EricTowers: It is easier to be connected in a coarser topology. – copper.hat Jul 09 '14 at 03:55

@copper.hat: Well that's true. Hmm... – Eric Towers Jul 09 '14 at 03:56

@EricTowers: However, there must be something wrong with my convex > path connected > connected reasoning since convex is independent of topology and if we choose the discrete topology, then every set (with at least two elements) is disconnected. – copper.hat Jul 09 '14 at 04:00

@EricTowers, I guess the mathexchange link you shared shows that $[0,M]^\infty$ is connected for the product topology. What distance function would that topology imply for the infinite case? – Juanito Jul 09 '14 at 04:00

@Juanito: In this case, since the function $f: [0,1] \to [0,M]^\infty$ given by $f(t) = t x_1+(1t) x_2$ (where $x_k \in [0,M]^\infty$) is continuous (with the metric $d$), it follows that $[0,M]^\infty$ is path connected and hence connected. – copper.hat Jul 09 '14 at 04:04

@copper.hat: You should post your half of the answer so we place Juanito in the position of having to decide which half to accept. :) – Eric Towers Jul 09 '14 at 04:09

I assume you meant $d(x,y) = \sup_t x_ty_t$? – copper.hat Jul 09 '14 at 04:45

1Related questions: http://math.stackexchange.com/questions/660418/whyislinftynotseparable, http://math.stackexchange.com/questions/97648/whyarelinftyandlinftynonseparablespaces (and maybe there are more posts on this site about separability of $\ell_\infty$). – Martin Sleziak Jul 09 '14 at 06:18
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(I am assuming that your distance is $d(x,y) = \sup_t x_ty_t$.)
Note that $X=[0,M]^\infty \subset l_\infty$ and $d(x,y) = \xy\_\infty$.
$X$ is not separable because the set $E=\{ x  x_k \in \{0,M\} \}$ is uncountable and the distance between any two distinct elements is $M$.
(To see why, suppose the space is separable, that is some $\{p_n\}$ that is dense in $X$. Take the uncountable collection of pairwise disjoint open sets $B(x,{M \over 2})$. Then each of the sets must contain at least one element of $\{p_n\}$, which is a contradiction since the latter set is countable.)
$X$ is connected because it is a convex subset of a topological vector space.
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What does distance have to do with having a countable dense subset? – Eric Towers Jul 09 '14 at 04:50

You are right to take that distance function. I just noticed that it was not clear in my question. – Juanito Jul 09 '14 at 04:56

Also, could you kindly explain some parts of your answer. Why do you use {0,M} in $\{ x  x_k \in \{0,M\} \}$, is it a typo? – Juanito Jul 09 '14 at 05:04

@Juanito: He means the set of points which have any sequence of $0$s and $M$s as coordinates. It is still unclear what this closed set (this space is T1) has to do with separability (i.e., having a countable dense subset). – Eric Towers Jul 09 '14 at 05:07

I mean all sequences $(x_1,x_2,....)$ where each $x_k$ is either $0$ or $M$. – copper.hat Jul 09 '14 at 05:08


I understand that the binary infinite elements you mentioned are uncountable and have a "distance" of M, but could you clarify the intermediate step one needs to imply that there cannot exist a countable dense subset? – Juanito Jul 09 '14 at 05:16

D'oh. I knew this. In finding an instance of this proof before your edit, I am reminded that we really should have had Juanito resolve whether he meant $[0,M]^\infty$ or $[0,M]^\omega$. You have nailed the $[0,M]^\omega$ version. – Eric Towers Jul 09 '14 at 05:17

@EricTowers: I have added an elaboration above. If a space has an uncountable collection of pairwise disjoint open sets, then it cannot be separable. – copper.hat Jul 09 '14 at 05:18

1@Juanito: Slightly longer version of this argument: http://mathprelims.wordpress.com/2008/07/16/linftyisnotseparable/ – Eric Towers Jul 09 '14 at 05:18

@EricTowers: What is (are?) the other interpretation(s) of $[0,1]^\infty$? (Box vs. product topologies?) – copper.hat Jul 09 '14 at 05:20

Munkres (and I've seen it elsewhere) gives $\mathbb{R}^{\omega}$ as the set of sequences of reals and $\mathbb{R}^{\infty}$ as the set of sequences of reals where only finitely many of the sequence members are nonzero. (Guess which one better matches the product topology...) – Eric Towers Jul 09 '14 at 05:22

I understand that I need to find uncountably many pairwise disjoint nonempty open sets, and argue that any dense set must have at least one of its points in any nonempty open subset of the space. – Juanito Jul 09 '14 at 05:24

@Juanito: That's correct, if the space was separable there would be a **countable** dense subset. If each of the uncountable open sets contains one (distinct) element of the countable set, then there is a contradiction. – copper.hat Jul 09 '14 at 05:33