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Let $A\subset X$ be connected. Then I want to conclude that $\bar A$ is connected. So I was trying to do the same thing as in the question of this post The closure of a connected set is connected

The thing is that I don't follow the answer, and in fact I think there is something wrong with it. So Can someone explain me why that answer is right? or provide a detail answer for this fact please?

In fact I don't follow because of @dunstin's comment, since I think he is right.

Thanks a lot in advance.

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user162343
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2 Answers2

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A space $X$ is connected iff every continuous map from $X$ into $\{0,1\}$ (with the discrete topology) is constant. (This is clear from the definitions; e.g. if such an $f$ were not constant, the inverse images of $\{0\}$ and $\{1\}$ would decompose $X$, etc.)

If $D \subseteq X$ is dense and two functions $f,g: X \rightarrow Y$, where $Y$ is Hausdorff, agree on $D$ (so $\forall x \in D: f(x) = g(x)$), then $f = g$ on $X$. This is a standard fact as well.

The combination of these two immediately implies your fact: suppose $A$ is connected and $A \subseteq B \subseteq \overline{A}$.

Suppose $f: B \rightarrow \{0,1\}$ is continuous. Then $f|A$ is also continuous, so $\forall x \in A: f(x) = i_0$ for some fixed $i_0 \in \{0,1\}$ ($f$ is constant on the connected set $A$). But then $f$ and $g$, the constant map to $i_0$, agree on the dense set $A$ (the closure of $A$ in $B$ is just the closure of $A$ in the whole space intersected with $B$, but by how $B$ lies in-between them, this just equals $B$), so they agree on $B$. So $f$ is constant and $B$ is connected.

Henno Brandsma
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  • Hello @Henno-Brandsma :), Thanks for your help but I haven't prove the function result and I don't know topology, Can you provide a more naive answer ? Thanks a lot. – user162343 Nov 08 '15 at 15:01
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A more direct answer using the definitions and the definition of subspace topology:

Suppose $A \subseteq B \subseteq \overline{A}$ are subsets of a space $X$ and suppose that $A$ is connected. I claim that $B$ is connected as well. (This is a bit more general, and is the same amount of work.)

So suppose $B = U \cup V$ where $U,V$ are disjoint open subsets of $B$. By the definition of the subspace topology, we can write $U = \hat{U} \cap B$, $V = \hat{V} \cap B$, where $\hat{U}, \hat{V}$ are open subsets of $X$.

Then clearly $A \subseteq B = U \cup V \subseteq \hat{U} \cup \hat{V}$, so $A = (\hat{U} \cap A) \cup (\hat{V} \cap A)$. Also $(\hat{U} \cap A) \cap (\hat{V} \cap A) \subseteq (\hat{U} \cap B) \cap (\hat{V} \cap B) = U \cap V = \emptyset$. As $\hat{U} \cap A$ and $\hat{V} \cap A$ are open subsets of $A$ (in the subspace topology), and as $A$ is connected the previous two facts imply that one of $\hat{U} \cap A$ and $\hat{V} \cap A$ is the whole of $A$ and the other is empty.

So assume wlog that $A = \hat{U} \cap A$, or equivalently $A \subseteq \hat{U}$. And so $\hat{V} \cap A = \emptyset$ or equivalently $A \subseteq X \setminus \hat{V}$.

Now we use that $\hat{V}$ is open, so its complement is closed and so $\overline{A} \subseteq \overline{X \setminus \hat{V}} = X \setminus \hat{V}$. This implies that $\overline{A} \cap \hat{V} = \emptyset$ and so a fortiori $B \cap \hat{V} = V$ is also empty. So we have shown one of $U$ and $V$ to be empty, and so $B$ is connected.

Henno Brandsma
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  • But the thing is that I don't know topology so I won't be able to use topology facts :( – user162343 Nov 08 '15 at 15:32
  • I don't understand. You talk about connectedness, that's topology. I only use the basic facts. – Henno Brandsma Nov 08 '15 at 15:44
  • @user162343 What do you mean by "I don't know topology?" You asked a question from topology, how can you not know topology? This answer is quite easy to follow. Please try to understand more carefully. – Babai Nov 08 '15 at 15:44
  • Ok, let me check it :), just give some time to get the idea please :) – user162343 Nov 08 '15 at 15:54