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Suppose $X$ is connected and locally path connected, then $X$ is path connected. Proof BWOC, let $Y$ is path component poper subset of $X$. Since $X$ is locally path connected then $Y$ is open and $X-Y$ is open. Then $Y$ is closed which is contradiction because $X$ is connected.

My question how can I prove it directly from the definition without depending on theorem.

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  • $\begingroup$ I upvote, not because of the question, but because of the BWOC proof. $\endgroup$ – user5280911 Jan 8 at 4:27
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Let $X$ be the space and fix $p \in X$. Let $C$ be the set of all points in $X$ which are path connected to $p$. Now, it is enough to show that $C$ is closed and open in $X$, to show that $C=X$ (given that $p \in C$ via a constant path, so $C$ is non-empty).

To show $C$ is open, let $c \in C$, then we can choose (by local path connectedness) an open subset $U$ containing $c$. For $u \in U$, $u$ is path connected to $c$ which is path connected to $p$, so by joining paths, we have that $u$ is path connected to $p$. In other words, $U \subset C$, so $C$ is open.

Look at the closure of $C$, namely $\bar{C}$. Let $d \in \bar{C}$, and choose, like above, an open path connected subset $V$ containing $d$. Note that $V \cap C \neq \phi$, because $V$ is open! Hence let $e \in V \cap C$, then $d$ is path connected to $e$ which is path connected to $p$ (because $e \in C$)! Hence $d$ is path connected to $p$ and $d \in C$, so $C=\bar{C}$, $C$ is closed.

Hence $C=X$, and $X$ is path connected.


EDIT : In the style of equivalence relations , we can create another proof. Indeed, define an equivalence relation on $X$ given by $p \sim q$ if there is a path connecting $p$ and $q$. It is easy to see that this is an equivalence relation : for reflexivity, use the constant path. For symmetry, use the reverse of a path, and for transitivity, use the concatenation of the two paths.

Finally, note that every equivalence class is open, because if I take $p$, then by local path connectedness, some neighbourhood of $p$ is path connected, but this includes $p$. Therefore, this entire neighbourhood is in the same equivalence class as $p$. The openness follows.

Now, the equivalence classes partition $X$ into disjoint open sets. Since $X$ is connected, this can't happen unless there's only one equivalence class, which is the whole of $X$. We are done!

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  • $\begingroup$ Do you mean that the open sets $U$ and $V$ in your answer are path-connected? $\endgroup$ – Oskar Henriksson Mar 8 '18 at 13:22
  • $\begingroup$ Yes, they must be path connected. $\endgroup$ – Teresa Lisbon Mar 10 '18 at 2:41
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    $\begingroup$ A quite obvious remark: The connectedness of $X$ enters the argument at the very end, when concluding from the fact that $C$ is open and closed, that $C=X$. $\endgroup$ – klirk Aug 24 '19 at 19:00
  • $\begingroup$ @klirk True, thank you for that. $\endgroup$ – Teresa Lisbon Aug 25 '19 at 7:15
  • $\begingroup$ @астонвіллаолофмэллбэрг, to show $C$ is closed could we argue in the same way as previously? $X \backslash C$ is opend for if we take $q \in X\backslash C$, there is an open set $U$ containing q which is path connected. $U\cap C = \emptyset$ because otherewise we can find $r$ in the intersection which is path connected to $p$ and path connected to q which is not possible. $\endgroup$ – roi_saumon Apr 27 '20 at 18:30
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I recommend looking up some of the equivalent definitions of connectedness, it is a useful exercise to prove the equivalences:

enter image description here

In particular characterization (6) is what you're looking for.

If I recall of the many definitions I found this equivalence the hardest to prove (one direction is straight-forward, the other a little tricky), it helped to draw a picture. Anyways, given for now the equivalence, the proof you're asking about is (hopefully) almost immediate. To begin, fix two points in X. Around each point in X you can take a path connected open set, this forms an open cover of X, which admits a finite subcover with the property that "adjacent" open sets are non-empty and still path-connected so you can glue together a path.

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  • $\begingroup$ If this theorem is taken from book, if yes , could you give the name of this book becuse may be will ever helpfuly in the connected $\endgroup$ – Gob Aug 2 '16 at 0:02
  • $\begingroup$ It's from a set of notes from an old professor of mine. The definition of connected given is "X is called connected if X is not the union of two disjoint non-empty open sets." The notes don't supply the proofs, we were expected to do so ourselves. $\endgroup$ – mb- Aug 2 '16 at 0:05
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Let $x\in X$, consider $C$ such that for every $y\in C$ there exists a path between $x\in Y$. Let's show that $C=X$, suppose $C\neq X$, remark that $C$ is open: $y\in C$, there exists $y\in U$ $U$ open subset such that for every $z\in U$ there is path $c$ between $z$ and $y$. Compose $c$ with the path between $x$ and $y$ to obtain a path between $x$ and $z$.

$C$ is closed, let $y\in \bar C$ the adherence of $C$, you have again an open subset $y\in V$ path connected $V\cap C$ is not empty, take $z\in V\cap C$ compose a path between x and z with a path between z and y.

$C$ is open and closed, $C=X$ since $X$ is connected.

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  • $\begingroup$ It is obvious now thank you $\endgroup$ – Gob Aug 2 '16 at 0:06

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