Show that each connected component of a topological space $X$ is connected.

Question

Show that each connected component of a topological space $X$ is connected.

Here is my proof

My Proof: Choose a connected component $$C = [x] = \{y \in X \mid \exists \text{ a connected set containing both $x$ and $y$}\}$$

Now for each $y \in C$ let $C_y$ be the connected set containing $x$ and $y$. We claim that $C = \bigcup_{y \in [x]}C_y$. To prove this claim pick $\alpha \in \bigcup_{y \in [x]}C_y$, then $\alpha \in C_y$ for some $y \in C$. Since $C_y$ is a connected set in $X$ containing both $\alpha$ and $x$, it follows that $\alpha \in C$ hence $\bigcup_{y \in [x]}C_y \subseteq C$.

Conversely pick $\beta \in C$, then by construction $\beta \in C_{\beta}$ which is the connected set containing both $x$ and $\beta$. Thus $\beta \in \bigcup_{y \in [x]}C_y$ and we have $C = \bigcup_{y \in [x]}C_y$.

Now since $\bigcup_{y \in [x]}C_y$ all contain $x$, their intersection is nonempty and since each $C_y$ is connected it follows that $\bigcup_{y \in [x]}C_y$ is connected. Thus $C$ is connected. $\square$

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First of is my proof correct? Can I improve it in any way? I'm not too happy about my construction of $\bigcup_{y \in [x]}C_y$, because I said

"for each $y \in C$ let $C_y$ be the connected set containing $x$ and $y$"

but there's no need for $C_y$ to be unique and that made matters a bit hairy when I tried to show that $C \subseteq \bigcup_{y \in [x]}C_y$

Is there a better way to prove this?

Answer 1

Some comments.

You pick up a connected component $C$, an element $x \in C$ and then provide a definition (by extension) of $C$ which is fine.

But then you use the theorem saying that a union of connected sets all containing the same point is connected. This is not necessary.

Better to come back to the definition of a connected set. Suppose that $C$ is the union of two (relative) disjoint open subsets $C_1,C_2$ with $x \in C_1$. Then for $x \neq y \in C$, it exists a connected set $C_y$ containing $y$ and included in $C$. Necessarily $y \in C_1$ as if not $C_y$ would be the disjoint union of the non empty (relative to $C_y$) open subsets $C_1 \cap C_y, C_2 \cap C_y$. Therefore $C_2 = \emptyset$, proving that $C$ is connected.

Answer 2

I don't think it matters too much that the sets $C_y$ may not be unique. The proof rests on the notion that a union of connected sets with common intersection is connected, which seems plausible (I haven't tried to prove it though). If that isn't an established proposition in your text though, I think it should be proved.

But the construction seems a little complicated. You might just appeal to the definition of connected. If $C = [x]$ were not connected, you could find disjoint sets $X$ and $Y$ such that $C \cap X \neq \varnothing$, $C \cap Y \neq \varnothing$, and $C \subseteq X \cup Y$. Choose an $x\in X \cap C$ and $y \in Y \cap C$, and a connected set $D$ containing $x$ and $y$. Can you argue to a contradiction from here?

Answer 3

This is just an expansion of mathcounterexamples.net's answer above.

Proof: Choose a connected component $C=[x]$. Suppose that $C$ is disconnected, then $C = C_1 \cup C_2$ where $C_1$ and $C_2$ are open in $C$ and disjoint and both are nonempty. Without loss of generality assume $x \in C_1$

We claim that $C_2 = \emptyset$. To prove this claim suppose that $C_2$ was non-empty, then pick $y \in C_2$, then by definition of a connected component (since $y, x \in C$ implies $y \sim x$) there exists a connected set $C_y$ containing both $y$ and $x$.

Note that $C_y \subseteq C$, to see why this is true, pick $\alpha \in C_y$, then by definition of the relation $\sim$, it follows that $\alpha \sim y$ (since $C_y$ is a connected set in $X$ containing both $\alpha$ and $y$) and since $y \sim x$ and the fact that equivalence relations are transitive it follows that $\alpha \sim x$ which implies that $\alpha \in [x] = C$. Hence $C_y \subseteq C$.

Now since $C_y$ contains both $x \in C_1$ and $y \in C_2$ and $C_y$ is connected and contained in $C$, it follows that $C_1 \cap C_y$ and $C_2 \cap C_y$ is open in $C_y$ and furthermore $C_1 \cap C_y \neq \emptyset$ (since both sets contain $x$) and $C_2 \cap C_y \neq \emptyset$ (since both sets contain $y$) and furthermore $(C_1 \cap C_y) \cup (C_2 \cap C_y) = C_y$ (since $C_1 \cup C_2 = C$ and $C_y \subseteq C$), thus $C_y$ is disconnected a contradiction. Thus $C$ must be connected. $\square$

Answer 4

Let $C$ be a connected component and let $x\in C$.

For every $y\in C$, choose a connected set $A_y$ that contains both $x$ and $y$. Then, by definition, $A_y\subseteq C$ and therefore $$ C=\bigcup_{y\in C}A_y $$

Theorem. If $(B_i)_{i\in I}$ is a family of connected subsets of the topological space $X$ such that $\bigcap_{i\in I} B_i\ne\emptyset$, then $\bigcup_{i\in I}B_i$ is connected.

Proof. Let $f\colon\bigcup_{i\in I}B_i\to\{0,1\}$ (discrete codomain) be continuous. Let $p\in\bigcap_{i\in I}B_i$ and let $q\in\bigcup_{i\in I}B_i$, say $q\in B_{i_0}$. Then $f(p)=f(q)$, because $f$ restricted to $B_{i_0}$ is continuous, hence constant. Therefore $f$ is constant. QED

Now note that $x\in\bigcap_{y\in C}A_y$.