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Consider the "infinite broom" $X$ pictured in figure below. Show that $X$ is weakly connected at $p$.

infinite broom

My attempt : I found the answer here.But im not satisfied with answer

My attempt : Consider $X$ to be a subspace of $\mathbb{R}^2$ with $p=(0,0) $ and each $a_n$ on the positive $x-$ axis .Let us assume $V$ as an open set in $\mathbb{R}^2$ and $U$ be an open neighbhood of $p$

By the definition of subspace topology $U= X \cap V $ is open/closed. This implies $V$ contains an open ball $B(p,a_n)$.

Now take connected subspace $Y$ of $X$ where $Y= U'\cup \{a_{n+1}\}$ because $U' \cap a_{n+1} \neq \emptyset$ where $U'=X \setminus B(p,a_{n+1})$

Therefore $Y \subset U$ and contains the neighbhorhood $X \cap B(p,a_{n+1})$ of $p$

Now by the Munkres book definition $X$ is weakly locally connected at $p$

A space $X$ is said to be weakly locally connected at $x$ if for every neighborhood $U$ of $x$, there is a connected subspace of $Y$ contained in $U$ that contains a neighborhood of $x.$

Note : Take $x=p$

Is my proof is correct or not ?

jasmine
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    I am more satisfied with the linked argument. You pick $V$ and $U$ and then say that $U=X\setminus V$. If the $V,U$ were chosen already, they could have been chosen without that property. The choice of $V$ must be made dependent on $U$. You probably want $X\cap V$ rather than $X\setminus V$. The $n$ in the next sentence hasn't been defined. – plop May 20 '21 at 15:01
  • @plop you are right it should be $X\cap V$ . But linked argument doesn't follow the Munkres definition . – jasmine May 20 '21 at 15:09
  • It does prove the same. They produce a connected subspace $Y$ that is contained in the given neighborhood $U$, contains $p$, and contains a neighborhood of $p$. The neighborhood of $p$ that it contains is the interior of $Y$. – plop May 20 '21 at 15:14
  • @plop that argument say that the subspace of $X$ including $p$ and every broom until the $n$th-one that mean it is a straight line from point $p$ to $a_n=[p,a_n]$ which is not open but closed set . we know that interior $[p,a_n] =(p,a_n)$ doesn't contains p i,e interior doesn't contains point $p$ – jasmine May 20 '21 at 15:20
  • When I read "broom" in that sentence I understood not just the segments $[a_{n+1},a_n]$, but also the other little segments that depart from $a_n$. – plop May 20 '21 at 15:24
  • okss @plop that mean subspace of $X$ including $p$ and every broom until the $n$th-one , $n>N$ are the intersection of closed half plane to the left side of $a_n $ and $U$.So this space is obviously closed set and connected but not open .Am i right ? – jasmine May 20 '21 at 15:38
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    Yes, the intersection of the closed half plane at $a_n$ and $X$. That intersection is a connected subspace of $X$, that contains $p$ in its interior, and $a_n$ was chosen such that it is included in $U$. – plop May 20 '21 at 15:49
  • oks, thank u @plop . Now now I've understood it – jasmine May 20 '21 at 15:51

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