This particular question was asked in my Topology quiz and I was unable to solve it and so I am asking for help here .
Question: Consider the following Subsets of $\mathbb{R}^2 : X_1 ={(x, sin(1/x) |0<x<1}, X_2 =[0,1]\times {0} , X_3 =${(0,1)} . Then ,
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$X_1 \cup X_2\cup X_3 $ is connected ;
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$X_1 \cup X_2 \cup X_3 $ is path connected;
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$ X_1 \cup X_2 \cup X_3 $ is not path connected , but $X_1 \cup X_2 $ is path connected ;
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$X_1\cup X_2$ is not path connected , but every open neighbourhood of a point in this set contains a smaller open neighbourhood which is path connected .
I have proved A to be true . I can visualize how the Diagram $X_1\cup X_2\cup X_3$ looks and intutively I think that it's path connected . ( Answer of question says that I am wrong )
But what difference does adding $X_3$ to $X_1 \cup X_2$ makes ? ( So , I am confused on option 2,3,4)
Actually , topology course was taught at my university but it was taken by a really terrible instructor who was not interested in teaching although I have self studied all of general topology from Wayne Patty's "Foundations Of Topology" .
It is my humble request to give a rigorious proof of the union of sets which is path connected and also give hint of what difference $X_3$ is making .
Answer :
1,3
I shall be really thankful for a detailed answer .
Best Answer
2,3)) Suppose to the contrary that a set $X=X_1\cup X_2\cup X_3$ is path-connected. Then there exists a continuous map $f(X)\to [0,1]$ such that $f(0)=(0,0)$ and $f(1)=(0,1)$. Put $X_-=\{(x,y)\in X:y\le 1/2\}$. Since the set $f^{-1}(X_-)$ contains $0$, it is non-empty. Since $X_-$ is a closed subset of $X$ and the map $f$ is continuous, a set $f^{-1}(X_-)$ is a closed subset of $[0,1]$, and so $f^{-1}(X_-)$ is compact. Thus a set $f^{-1}(X_-)$ contains its supremum $T$. Since $f(1)\not\in X_-$, $T<1$. It is easy to see that a set $X\setminus X_-$ splits into connected components which are the arcs of the graph and the set $X_3$. Since a set $(T,1]$ is connected, its continuous image $f((T,1])$ is connected too. Since $(0,1)\in f((T,1])\subset X\setminus X_- $, the only possibility to keep $f((T,1])$ connected is to have $ f((T,1]=\{(0,1)\}$. By the continuity of $f$, the set $f^{-1}(0,1)$ is closed, so it contains $T$. Thus $f(T)=(0,1)$, a contradiction with $f(T)\in X_-$.
3,4)) The set $X_1\cup X_2$ is path-connected, being a union of two intersecting path-connected sets (a segment and a graph of a continuous function on an interval).