Prove that every path-component is a path-connected space

general-topologysolution-verification

Prove that, in any topological space, every path-component is a path-connected space.

Definitions:

  1. The path-component in $X$ of $x$, $P_X(x)$, is the union of all path-connected subsets of $X$ which contain $x$.
  2. A path-connected subset is one where for each pair of distinct points $a$, $b$ in $X$, there exists a continuous mapping $f : (0,1) \rightarrow (X,\tau)$ such that $f(0)=a$ and $f(1)=b$.

My attempt:
$P_X(a)=\bigcup A_i$ where $A_i$ is path-connected and $x \in A_i$, and $i \in I$, $I$ being some index set.
Since $A_i$ is path connected, there exists $f_i : (0,1) \rightarrow (A_i,\tau)$ where $f_i(0)=a$ and $f_i(1)=b$. Also, there is some $p_i\in(0,1)$ such that $f_i(p)=x$ because $x$ is in each $A_i$.

I'm not sure of how I should proceed after this.
I know that I need to find $g : (0,1) \rightarrow (P_X(a),\tau)$ such that $g(0)=min(a_i)$ and $g(1)=max(b_i)$. I also can see intuitively that if the subsets all contain $x$ and are path-connected, their union should be path-connected.

Attempting to continue:
Let $f_j: (0,1) \rightarrow (A_j,\tau)$ such that $f_j(0)=a_{min}$,
and $f_k: (0,1) \rightarrow (A_k,\tau)$ such that $f_k(1)=b_{max}$.
Then, $P_X(a)=\bigcup A_i$ = $(a_{min},b_{max})$ which is path-connected?

Best Answer

Your solution is very hard to understand.

  • You write $P_X(a)$ and then $x \in A_i$. What is $x$?
  • "Since $A_i$ is path connected there exists $f_i$ with $f_i(0) = a$ and $f_i(1) = b$." No - you have no guarantee that $a$ and $b$ are in $A_i$.
  • "There is some $p_i \in (0,1)$ such that $f_i(p) = x$" - What is $p$? What is $x$? Did you mean $P_X(x)$? Do you mean $f_i(p_i)$? If you do, why must $x$ be on the path you chose from $a$ to $b$ (if one exists)? If you know $x \in A_i$, and $f_i$ is a path in $A_i$, why does it follow that $x$ is on that path?
  • "I know I need to find ... with $g(0) = \min a_i$ and $g(1) = \max b_i$." What are $a_i$ and $b_i$? You only have $a$ and $b$!

What you actually need to prove is:

Given a path component $P_X(x)$, and a pair of distinct points $a,b \in P_X(x)$, there exists a path $f: (0,1) \to P_X(x)$ such that $f(0) = a$ and $f(1) = b$.

By definition, there are paths $g_a$ and $g_b$ which connect $x$ to $a$ and $x$ to $b$, respectively (although you should explain how). Can you use them to construct a path from $a$ to $b$?

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