I'm currently studying Springer's An Introduction to Algebraic Topology, and after the definition of the cone $CX$ over a topological space $X$, he remarks that the quotient topology on $CX$ for when $X$ is not compact Hausdorff may "have more open sets than expected". He gives an example in a footnote:
Let X be the set of positive integers regarded as points on the x-axis in $\mathbb{R}^2$; let C'X denote the subspace of $\mathbb{R}^2$ obtained by joining each $(n, 0)\in X$ to $v=(0, 1)$ with a line segment. There is a continuous bijection $CX\to C'X$, but $CX$ is not homeomorphic to $C'X$.
I'm struggling to see how these two spaces are not homeomorphic. Springer's remark would suggest that there is an open set in $CX$ for which the image under the obvious map to $C'X$ is not open, but I can't find such a set.
For reference, the definition of the cone $CX$ given is:
Definition. If X is a space, define an equivalence relation on $X\times I$ by $(x, t)\sim (x',t')$ if $t=t'=1$. Denote the equivalence class of $(x, t)$ by $[x, t]$. The cone over $X$, denoted by $CX$, is the quotient space $X\times I/\sim$.
Take the set consisting of points within $1/n$ of $(0,1)$ on the segment joining $(0,1)$ with $(n, 0)$. The preimage is open in $X \times [0,1]$ so this is open in $CX$. But for any $\epsilon > 0$ this misses points within $\epsilon$ of $(0, 1)$ so is not open in the subspace topology on $C'X$.