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I have some confusion on my understanding of connectedness of subsets of $\mathbb{R}^2$. For example I was given a practice problem to explain why the comb space $S=[0,1] \times \{0\} \cup \{(0,1)\} \cup (\cup_{n=1}^{\infty}\{\frac{1}{n}\} \times [0,1])$ is connected, and the answer says "Since any ball around the point $(0,1)$ contains some points on the spikes(near $(0,1)$ , we cannot find an open set separating $(0,1)$ from the rest of $S$".

Now I am seeing a practice problem in topology by Franzosa which says the union of the two subsets $y=0$ and $y=e^x$ is not connected. But then I think about the explanation given for the comb space, and it seems I can always find an open ball of any given radius around some point arbitrarily far down the negative $x-$axis that will intersect $y=e^x$.I do not need rigorous answers for this, I just need a conceptual understanding. In what way do these situations differ, so that the comb space is connected, but not the latter?

Also, what would be the best way(without proof) to determine subsets of $\mathbb{R}^2$ are connected or not? Path connected?

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    $\begingroup$ No point on the graphs of either $y=0$ or $y=e^x$ is itself arbitrarily far down the the negative $x$-axis. You pick a point, and then the radius of the open circle may depend on the point. The point $(0,1) \in S$ would be analogous to a point at $x=-\infty$ on the graphs, but there is no such point (which is why the analogy is improper). $\endgroup$ Mar 17 '21 at 5:11
  • $\begingroup$ "I can always find an open ball of any given radius around some point arbitrarily far down the negative $x$−axis that will intersect $y=e^x$." This statement is unclear, what do you mean by "any given radius?" Balls with large radius will intersect $y=e^x$ BUT balls of small radius will NOT. Given any radius $r$ you could find $x_r$ down the $x$-axis so that the ball with radius $r$ and center $x_r$ does intersect $y=e^x$. BUT you could then find a smaller radius $r'<r$ such that the ball with center $x_r$ and radius $r'$ does NOT intersect $y=e^x$. $\endgroup$
    – Mirko
    Mar 17 '21 at 5:24
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"Since any ball around the point $(0,1)$ contains some points on the spikes near $(0,1)$, we cannot find an open set separating $(0,1)$ from the rest of $S$."

If that should be the only argument to prove the connectivity of $S$, then it is sheer nonsense. To see this, consider $$S' = \{(0,1)\} \cup (\bigcup_{n=1}^{\infty}\{\frac{1}{n}\} \times [0,1]) .$$ This space is certainly not connected (each of the sets $\{\frac{1}{n}\} \times [0,1])$ is open and closed in $S'$), but we can correctly argue

"Since any ball around the point $(0,1)$ contains some points on the spikes near $(0,1)$, we cannot find an open set separating $(0,1)$ from the rest of $S'$."

To give a complete proof, observe that the space $$T = [0,1] \times \{0\} \cup (\bigcup_{n=1}^{\infty}\{\frac{1}{n}\} \times [0,1])$$ is path connected, hence connected. Now assume that $S = U \cup V$ with disjoint nonempty open subsets $U, V$ of $S$. W.l.o.g. $(1,0) \in U$. But now $(1,0) \in B_r(1,0) \cap S \subset U$ for some $r > 0$. Your argument says that $B_r(1,0)$ must contain a point of $T$, thus $U \cap T \ne \emptyset$. But then $T = U' \cup V$ with $U' = U \cap T$. The sets $U',V$ are disjoint nonempty open subsets of $T$, a contradiction since $T$ is connected.

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