Connected and locally path connected

Question

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.

Answer 1

I recommend looking up some of the equivalent definitions of connectedness, it is a useful exercise to prove the equivalences:

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.

Answer 2

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.

Answer 3

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!