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Let $(M,d)$ be a metric space, which is path-connected. Fix $p \in M$. Now, $\forall q \in M-\{p\}$, $\exists f_q:[a,b] \rightarrow M$ with $f_q(a)=p$ and $f_q(b)=q$, and $f_q$ is continuous. Call each such $[a,b]$ a domain interval, which is not unique.

Let $\{I_q| q \in M-\{p\}\}$ be a collection of such domain intervals (note that such a collection is not unique). Now, using the fact that $I_q$ is connected for each $q \in M-\{p\}$, and the fact that the functions $f_q$ are continuous, $f_q(I_q)$ must be connected as well. Moreover, the intersection of $f_q(I_q)$ over $q \in M-\{p\}$ is non-empty, as $p$ must belong to this intersection. Hence, their union must be connected. Now, $\forall q \in M-\{p\}$, $f_q(I_q) \subset M$ by definition of $f_q$. Also, $\forall q \in M-\{p\}$, $q \in f_q(I_q)$, again by definition. Of course, $p \in f_q(I_q )$ $\forall q \in M-\{p\}$ as well. Thus, $\bigcup\limits_{q \in M-\{p\}} f_q(I_q) = M$ and $M$ is connected $\blacksquare$

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  • $\begingroup$ Added to the proof @drhab $\endgroup$ – Student Aug 29 '20 at 14:13
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    $\begingroup$ I think your proof is okay but only valid if you first proved that a union of connected spaces that have an element in common is again connected. This can be avoided as is shown in the answer to your question. Further it is not necessary to give the intervals an index $q$. For every $q\in M$ you can just use interval $[0,1]$. Indexing the functions is enough. $\endgroup$ – drhab Aug 29 '20 at 14:34
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    $\begingroup$ Each interval is a domain interval. The concept does not make much sense. $\endgroup$ – Paul Frost Aug 29 '20 at 22:17
  • $\begingroup$ @drhab I agree, but the book I'm using proves that "union of connected spaces that have an element in common is again connected" first, hence I proceeded this way. Thanks for your input. $\endgroup$ – Student Aug 31 '20 at 10:24
  • $\begingroup$ @PaulFrost I agree, the set of domain intervals is not unique, but we can choose any intervals we please and put in there, even just $[0,1]$. $\endgroup$ – Student Aug 31 '20 at 10:24
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$X$ is connected iff for each $x,y \in X$ there exists a connected subspace $C(x,y)\subseteq X$ such that $x,y \in C(x,y)$.

Left to right is trivial, we can take $C(x,y)=X$ always. Right to left: suppose $X$ is not connected while the right hand side holds, write $X=A \cup B$ where $A,B$ are disjoint, non-empty and both open. Pick $a \in A, b \in B$ and for the $C(a,b)$ that exists, note that $C(a,b) = (A \cap C(a,b)) \cup (B \cap C(a,b))$, so that $C(a,b)$ is not connected, which is a contradiction. So $X$ is connected.

(I could also have shown, as in your proposed proof that (for some fixed $p \in X$), that $$X= \bigcup\{ C(p,a): a \in X\}$$ which is a union of connected subspaces that all intersect at $p$ and so is connected, but this needs an extra theorem, while the previous only needs the definition of connectedness).

Now, note that the right hand side is easily satisfied for a path-connected space: if $f:[a,b] \to X$ is a path from $x$ to $y$, use $C(x,y)=f[[a,b]]$, which is connected as $f$ is continuous and $[a,b]$ is always connected.

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Correct.

Another way of proving it is to prove the false converse implications. So not connected implies not path connected. Easy if you assume that $M$ is path connected. You show that, since $M$ is not connected there are points belonging to different connected components that are not "reachable" by continuous path (as you said the image would be connected) and thus this is an absurdum.

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Suppose that $M$ is not connected.

Then non-empty open disjoint sets $A,B$ exist with $A\cup B=M$.

For every continuous $f:[a,b]\to M$ the image $f([a,b])$ is connected, hence must be a subset of $A$ or a subset of $B$.

So if $p\in A$ and $q\in B$ a path connecting $p$ and $q$ does not exists.

We conclude that $M$ is not path-connected.

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