If every nbd of $x$ also contains $y$, does there exist a path connecting $x$ and $y$?

Question

If $x,y \in X$ are two points in a topological space, and every neighborhood of $x$ also contains $y$, are $x$ and $y$ connected by a path?

I don't really know how or where to start. If $\{ U_i\}_{i \in I}$ are the described open sets, doesn't the constant sequence $(x)_{i \in I}$ converge to $y$? A sequence converges if and only if it is eventually constant, and here it is always constant. But isn't it necessary for $I$ to be infinite? In any case, I can't translate this into paths. Thanks in advance for any help.

Answer 1

Hint: It's not a counterexample, but it'll at least get you thinking in the right direction....take a look at the "line with two origins" to see why your convergence argument isn't valid. (Well...it's valid...but the limit you get is $x$ rather than $y$.)

Note: a sequence won't help anyhow, because you need to build a path, not just a sequence of points. Otherwise the points $0$ and $1$ in $\Bbb Q$ would be "connected".

Answer 2

Here are two observations that may help.

1. Consider the two-point subspace $A=\{x,y\}$ of $X$. What are the possible subspace topologies that $A$ can inherit?

2. You can define paths in finite spaces. For example, if $B=\{x,y,z\}$ is a space with topology $\{B,\{x\},\{z\},\{x,z\},\emptyset\}$, the function $p:[0,1]\to B$, $p([0,1/2))=x$, $p(\{1/2\})=y$, $p((1/2,1])=z$ is a path since it satisfies the definition of continuity.