Dugundji's identifiaction topology example 
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.
 A: 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 )

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.
A: 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 $CX$.

*

*$V$ 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 $1/n < r$.


*$f^{-1}(V)$ 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 $CX$.
