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Let $X=\{(n,0) \mid n \in \mathbb{Z}^+\}$ and let $CX = (X \times I)/(X \times \{1\})$. Let $TX$ be the subspace of $\mathbb{R}^2$ obtained by joining each $x \in X$ to $w_0 =(0,1)$ by a segment. Show that the spaces $CX$ and $TX$ are not homeomorphic.

I am reading Dugundji's topology about the identifiaction spaces and this is the first example given. He says that the space are not homeomorphic because if one lets $V_n = \text{ all points on the segment $w_0$ to $(n,0)$ within $1/n$ of $w_0$}$ and $V = \bigcup_n V_n$, then $V$ is open in $CX$, but not in $TX$.

Why is this true? Is there a picture which explains this I don't understand what the $V_n$ is supposed to represent.

Edit: The linked duplicate uses compactness which comes after the chapter I am reading so I would want to do find out how this is proved without compactness arugments.

Ricardo
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    You changed notation: What Dugundji denotes as $TX$ and $CX$, respectively, is denoted in your question by $CX$ and $TX$, respectively. Concerning $CX$ your variant is standard. But then you have to write "$V$ is open in $CX$, but not in $TX$". – Paul Frost Oct 08 '22 at 11:06
  • @PaulFrost Yes I did just because I thought that $CX$ was understood better as the quotient space for people who haven't got access to the book. – Ricardo Oct 08 '22 at 11:20

2 Answers2

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Open subsets of $\mathbb{R}^n$ arise from open balls. So take an open neighbourhood $V$ of $w_0$ in $TX$. This means there is an open ball of radius say $r$ around $w_0$ inside $V$. This means that for each $x\in X$ our $V$ contains an interval starting at $w_0$, towards $x$ of length $r$. We cannot have a situation in $\mathbb{R}^2$ that for each $x\in X$ we take smaller, and smaller (converging to length $0$) intervals from $w_0$ to $x$, and we compose all those intervals into an open subset. Here's the picture (and forgive me my drawing skills :D )

enter image description here

Note that those $I_n$ (which is the same as what Dugundji denotes by $V_n$, I just changed it to $I$, since these are intervals) have smaller and smaller lengths, converging to $0$. Such union cannot be open. By the earlier argument $\{length(I_n)\}_{n=1}^{\infty}$ has to be bounded from below by a positive real.

But such phenomenon does happen in $CX$. For each $(n,0)\in X$ you take interval $I_n=\{(n,0)\}\times (1-1/n,1]$. The union $\bigcup I_n$ is open in $CX$, but not in $TX$.

This is of course informal, those intervals are not even subsets of $TX$, plus even if they were, it doesn't mean there is no homeomorphism. But that's the intuition behind the argument.

freakish
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  • Thank you for the great explanation. I think I don't understand the explanation from Dugundji because I don't quite understand the sentence you wrote that "We cannot have a situation in $\mathbb{R}^2$ that for each $x \in X$ we take smaller, and smaller (converging to length $0$) intervals from $w_0$ to $x$, and we compose all those intervals into an open subset." Why cant we have this? Isn't it not an union of open sets which should also be open? – Ricardo Oct 08 '22 at 10:02
  • @Ricardo none of the intervals $I_n$ is open in $TX$. These are not open in $CX$ as well. They fail to be open, because neighbourhoods around $w_0$ have to contain pieces of other intervals. However the union of all $I_n$ is open in $CX$, but not in $TX$. – freakish Oct 08 '22 at 10:07
  • But aren't we considering open balls which are open? I understood is as follows. The $V_n$'s that Dugundji is considering are $V_n = B(w_0, 1/n) \cap I_n$ and then he is unioning over all these $V_n$'s, but aren't these open? – Ricardo Oct 08 '22 at 10:12
  • @Ricardo Dugundjis $V_n$ and my $I_n$ are literally the same thing: an interval (line segment) from $w_0$ to $(n,0)$ of length $1/n$. Sorry for the naming confusion. No, these are not open. Again, have a look at open balls around $w_0$. No open ball is fully contained in $I_n$ because each open ball intersects each $I_m$. – freakish Oct 08 '22 at 10:23
  • I think I got the idea. Any open ball around $w_0$ is filled with "disjoint" intervals $I_j$ etc for large enough $n$. So in $TX$ we are concerned with the subspace topology so we intersect this ball with $TX$ and end up with a disjoint union of multiple intervals. And this is not open since it's an union over closed sets? I think this last part is not true as arbitary union of closed sets could be open? @Freakish – Ricardo Oct 08 '22 at 10:27
  • @Ricardo yes, union of closed sets can be open. That's not the argument here (besides $I_n$ is not closed, the other end is missing). The argument is that lengths of those intervals converge to $0$, while open balls have positive radius. Meaning those lengths have to be bounded from below. Alternatively the complement of that union is not closed. Because you can take sequence convergent to $w_0$ fully contained in the complement. – freakish Oct 08 '22 at 10:30
  • Sorry for not understanding this, but what does the fact that the balls have a positive radius have to do with this? The lengths of $I_n$'s converge to $0$ so I would not have any issue putting an open ball around them? Few comments up you stated that no open ball is fully contained in $I_n$ because the ball intersects $I_m$, but how are we even trying to put open balls inside intervals in the first place? That is very counterintuitive. @freakish – Ricardo Oct 08 '22 at 10:44
  • @Ricardo I'm talking about open ball around $w_0$ in the union of all intervals $V$. Each $I_n$ is not open, but that is not that important, we are asking whether $V$ is open, not each $I_n$. And what does it mean for a set to be open? It means that each point in $V$ is contained with some ball around it. Ball in $TX$. But is there an open ball around $w_0$ fully contained in $V$? No, because balls have positive radius, while those intervals are smaller and smaller. – freakish Oct 08 '22 at 11:02
  • Okay. I was thinking this in the wrong way. The open balls in $TX$ are not "balls" they are intersections of the balls in $\Bbb R^2$ with the intervals or distinct points in $TX$. So they are themselves an intervals or singletons. So asking wheter $V$ is open is the same thing than asking wheter there is a basis set in the base of $TX$ which is contained in $V$. The base of $TX$ is ${ B \cap TX \mid B \text{ open ball in } \Bbb R^2}$. – Ricardo Oct 08 '22 at 11:16
  • Yes, sorry, I incorrectly assumed this is clear. – freakish Oct 08 '22 at 11:34
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The proofs given in the answers to Equivalent definition $\text{Cone}(K)$ and Cones over noncompact spaces are not unions of paths are more general, but let us follow Dugundji's advice.

First let us note that Dugundji is imprecise when he claims that the spaces $CX$ and $TX$ are not homeomorphic. What he really means becomes clear if we look at his sketch: There is a canonical map $f' : X \times I \to TX$ given by $f'(x_n,t) = tw_0 + (1-t)x_n$. Here $x_n = (n,0)$. Since $f'(X \times \{1\}) = \{w_0\}$, we get a continuous bijection $$f : CX \to TX .$$

Dugundji means that $f$ is not a homeomorphism, otherwise his sketch does not make any sense.

Dugundji defines $V_n = \{ y \in L(x_n, w_0) \mid \lVert w_0 - y \rVert < 1/n \}$, where $L(a,b)$ is the line segment connecting $a, b \in \mathbb R^2$. Writing $y = tw_0 + (1-t)x_n$ we get $y \in V_n$ iff $$\lVert w_0 - tw_0 - (1-t)x_n \rVert = \lVert (1-t_n)(w_0 - x_n) \rVert = (1-t_n)\lVert w_0 -x_n \rVert < 1/n ,$$ i.e. $t > t_n :=1 - 1/n\lVert w_0 -x_n \rVert$. Of course $V = \bigcup_n V_n$ is not a subset of $CX$, so Dugundji certainly means that $V$ is not open in $TX$, but $f^{-1}(V)$ is open in $CX$.

  1. $V$ is not open in $TX$.

If it were open, then it would contain some set $TX \cap B(w_0;r)$ with $r > 0$. Hence for each $n$ the set $V_n$ would contain all $y \in L(x_n,w_0)$ with $\lVert w_0 - y \rVert < r$. But this is false for $1/n < r$.

  1. $f^{-1}(V)$ is open in $CX$.

Let $p : X \times I \to CX$ denote the quotient map. The set $W = \bigcup_n \{ x_n \} \times (t_n,1]$ is open in $X \times I$ because $X$ is discrete, thus $V = p(W)$ is open in $CX$ since $p^{-1}(p(W)) = W$. But clearly $f^{-1}(V) = p(W)$.

Remark:

If we want to show that $CX$ and $TX$ are not homeomorphic, we cannot pick a special map and show that it is no homeomorphism. We need some additional argument. A general result was proved in the above two links.

Here is a variant adapted to the present case.

Assume that there exists a homeomorphism $h : CX \to TX$. Let $*$ denote the tip of $CX$ given by $\{ * \} = p(X \times \{1\})$.

We must have $h(*) = w_0$. In fact, since $CX \setminus \{*\}$ has infinitely many path components. Thus $TX \setminus \{y\}$ must have infinitely many path components for $y = h(*)$. The only point with this property is $y = w_0$. We conclude that $h$ maps each "line segment" $L'(x_n) = p(\{x_n\} \times I)$ homeomorphically onto some $L(x_{\phi(n)},w_0)$, where $\phi : \mathbb N \to \mathbb N$ is a bijection.

The above set $V$ is not open in $TX$, but we can easily show as above that $h^{-1}(V)$ is open in $CX$.

Paul Frost
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  • This answer is gold! Few questions I have. Firstly what's the point of defining $t_n$ on the go here? Secondly what do you mean by $\lVert w_0 - y \rVert < r$ being false for $1/n < r$? @PaulFrost – Ricardo Oct 08 '22 at 20:27
  • @Ricardo I showed that a point $y = tw_0 + (1-t)x_n)$ is contained in $B(w_0;1/n)$ iff $t > t_n$. This allows to compute $f^{-1}(V)$ in 2. and to express it via the $t_n$. In 1. I showed that if $V$ would contain some $TX \cap B(w_0;r)$, then $V_n$ would contain all points $y = tw_0 + (1-t)x_n)$ with $\lVert w_0 - y \rVert < r$. But by definition it only contains the points with $\lVert w_0 - y \rVert < 1/n$. Therefore, if $1/n < r$, then it is false that $V_n$ all points $y$ with $\lVert w_0 - y \rVert < r$. – Paul Frost Oct 09 '22 at 08:00