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Question. Prove that the closed unit square $S=\{x$ in $\Bbb R^2 : 0 \le x_1 \le 1, 0 \le x_2 \le1 \}$ is connected.

I understood how to prove unit interval is connected and I am trying to extend to unit square case. I imitated the procedure to prove the question above.

This is my attempt:

Suppose that $ [0,1] \times [0,1] \subset U \cup V$ where $U$ and $V$ are disjoint, opens in $\Bbb R^n$ . Without loss of generality, we may assume that $ (x_1, x_2)=(0,0) \in U.$ Then to complete the proof, it suffices to show that $ [0,1] \times [0,1] \subset U$

Step 1: Since $ (x_1, x_2)=(0,0) \in U$ and $U$ is open, there is $\varepsilon $ with $0<\varepsilon<1$ such that $N((0,0), \sqrt 2 \varepsilon) \cap S \subset U$ , implying that $[0,\varepsilon) \times [0,\varepsilon) \subset U$. Then the set $T=\{x\in [0,1] : [0,x) \times [0,x) \subset U\}$ is nonempty and bounded. Therefore, there exists $\mu=supT$. Note that $0<\varepsilon \le \mu \le 1.$

Step 2: We will show that $[0,\mu) \times [0,\mu) \subset U$. $y\in [0,\mu) \Rightarrow$ There exist $x \in T$ such that $y<x \le \mu \Rightarrow y \in [0,x) \Rightarrow x \in [0,\mu) \times [0,\mu) \subset [0,x) \times [0,x) \subset U$

Step 3: We will show that $[0,\mu] \times [0,\mu] \subset U$. To prove this, it suffices to show that $(x_1, x_2)=(\mu,0) \in U.$ Suppose to the contrary that $(x_1, x_2)=(\mu,0) \in V.$ Since V is open, there is $\varepsilon>0$ such that $N((\mu,0), \varepsilon) \cap S \subset V$.

I am stuck here. Am I doing right? Is there any simpler version of proof? I am a beginner so any kind of help would be appreciated.

user320317
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2 Answers2

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With respect to your original proof: I agree with Step 1, and I agree that Step 3 follows from Step 2; I haven't checked Step 2. Once you've got Step 3, you've got a contradiction unless $\mu = 1$: since $U$ is open, there is an open neighbourhood of $(\mu, \mu)$ in $U$, and this open neighbourhood isn't allowed to contain any points to the top-right of $(\mu, \mu)$ because $\mu$ is a sup.

There is a simpler version if you're allowed access to "path connected implies connected", because $[0,1]^2$ is obviously path-connected (indeed, convex).

Alternatively, you can use that the product of connected spaces is connected.

Patrick Stevens
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    Even if you can't directly use "path connected implies connected" as a lemma, you can recite the trivial proof. The higher-dimensional case can be reduced to the one-dimensional case by taking a line segment between a point in one piece of a hypothetical separation and a point in the other piece of the separation. The intersection of the pieces of the separation with the line segment will then give a separation of the line segment, which we know is impossible. This works in any topological vector space over the reals. – Carl Mummert May 18 '16 at 15:06
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Yes there is a simple version of the proof. You have already proof that $I$ the unit interval is connected. Then, if you check that product of connected space is also connected you have finished since the unit square is $I\times I$.

InsideOut
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