# Topology question about a special subset in $\mathbb R^2$

Problem Statement:

Let $$X = (\bigcup \limits_{n \in \mathbb N} \{\frac{1}{n}\} \times [0,1] ) \cup \{(0,0),(0,1)\}$$ have a subspace topology as a subspace of $$\mathbb R^2$$. For any separation $$U$$ and $$V$$ of $$X$$, if $$(0, 0) \in U$$, then $$(0, 1) \in U$$ as well.

My attempt:

By the result from Munkres, if $$U$$ and $$V$$ are a separation of $$X$$ and $$Y$$ is a connected subspace of $$X$$, then $$Y$$ is completely contained in either $$U$$ or $$V$$. Hence, to show that $$(0, 0) \in U$$ would imply $$(0, 1) \in U$$, it suffices to show that there exists some connected subspace of $$X$$ that contains both $$(0, 0)$$ and $$(0, 1)$$.

From here, I am having trouble finding some connected subspace of $$X$$ that contains both points.

Your solution will not work. Any neighbourhood $$W$$ of $$X$$ which contains both $$(0,1)$$ and $$(0,0)$$ can be disconnected either by the open sets $$\{(x,y)|y>\frac12\}$$ and $$\{(x,y)|y<\frac12\}$$ or if there exists $$n\in\mathbb{N}$$ with $$(\frac1n,\frac12)\in W$$, by the open sets $$\{(x,y)|x<\frac1n\}$$ and $$\{(x,y)|x>\frac1{n+1}\}$$.

Instead you can use the following argument:

If $$U$$ and $$V$$ separate $$X$$, then each vertical line $$L_n=\{\frac1n\}\times [0,1]$$ is connected, so by the result you mentioned completely contained in $$U$$ or $$V$$.

Any neighbourhood $$U$$ of $$(0,0)$$ will intersect all vertical lines $$L_n$$ for all $$n>m_1$$ for some $$m_1$$.

Similarly, any neighbourhood $$V$$ of $$(0,1)$$ will intersect all vertical lines $$L_n$$ for all $$n>m_m$$ for some $$m_2$$.

Thus if $$U,V$$ separate $$X$$, they will each contain all $$L_n$$ for $$n>\max(m_1,m_2)$$, yielding the desired contradiction.

As far as I can see, there is no connected subset of $$X$$ that contains both $$(0,0)$$ and $$(0,1)$$. However, you're on the right track with the idea that...

...if $$U$$ and $$V$$ are a separation of $$X$$ and $$Y$$ is a connected subspace of $$X$$, then $$Y$$ is completely contained in either $$U$$ or $$V$$.

Here's my proof idea. Let $$U$$ and $$V$$ form a separation of $$X$$ and suppose $$(0,0)\in U$$. As $$U$$ is open, there is a ball around $$(0,0)$$ contained in $$U$$. This ball must intersect all of the lines $$\{1/n\}\times[0,1]$$ for all $$n>n_0$$. By what you said, all these lines must be contained in $$U$$. Then since $$U$$ is closed, it must contain its limit points, and clearly the point $$(0,1)$$ is a limit point of the union of the lines.

There is no connected subset of $$X$$ containing $$(0,0)$$ and $$(0,1)$$.

Suppose $$(0,0) \in U$$ and $$(0,1) \in V$$. There exists $$N$$ such that $$(\frac 1 N, 0) \in U$$ and $$(\frac 1N, 1) \in V$$. Let $$U_1= U \cap (\{\frac 1 N\} \times [0,1])$$ and $$V_1= V \cap (\{\frac 1 N\} \times [0,1])$$. This gives a separation of the line segment $$\{\frac 1 N\} \times [0,1])$$ which is a contradiction.