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Every connected metric space with at least two points is uncountable.

This is baby Rudin exercise 2.19(d). My argument is the following:

Let $X$ be a connected metric space with at least two points.

If $A\subset X$, and $A\not=\varnothing$, then $A$ clopen $\implies$ $A=X$ by connectedness.

So my plan is to show for a non-empty countable open subset $A$, not to be closed. (for the case "non-empty countable closed $\implies$ not open", I think it is the contrapositive statement.) Then every non-empty countable open subset would be proper, which makes $X$ uncountable (for otherwise, $X$ would be a proper subset of $X$. for not open subsets, they are proper because $X$ is clopen. for "closed $\implies$ not open" subsets and not closed subsets follow the same logic, thus all proper). Basically I'm trying to mimic the Cantor's diagonal process except the diagonal part.

Now, $A$ countable $\implies A=\{a_n:n\in \mathbb{N}\}$.

Define $A_i=A\setminus \{a_1,a_2,...,a_i\}$ so that $A_1 = A\setminus\{a_1\},\ A_2=A\setminus\{a_1,a_2\},...$

Note that each $A_i$ is non-empty, open. (since $\{a_1,a_2,...,a_i\}$ is closed for each $i$, and $A$ is open.) (any finite set is closed because it has no limit points. so $A$ finite $\implies \overline{A}=A\cup A'=A\cup \varnothing=A$)

Now pick $b_i\in A_i$ for each $i$ such that $b_1\in A_1$, and $b_{i+1}\in A_{i+1}\cap B_{1/i}(b_i)$, where $B_{1/i}(b_i)=\{x\in X: d(x,b_i)<1/i\}$, an open ball centered at $b_i$ with radius $1/i$.

So the tail of a sequence $\{b_n\}$ shrinks. The existence of $b_{i+1}$ is guaranteed because both $A_{i+1}$, $B_{1/i}(b_i)$ are open, so is the intersection of them. Hence there exists $r>0$ such that $B_r(b_i)\subset A_{i+1}\cap B_{1/i}(b_i)$. And $B_r(b_i)\setminus\{b_i\}\not=\varnothing$ because otherwise $B_r(b_i)=\{b_i\}$ which is closed and open, but only non-empty clopen subset of $X$ is $X$ provided that it's a connected space, and $X$ has at least two elements from the assumption, thus it leads to a contradiction.

Finally, if $b_n\to l\in X$, then $l\not\in A$, because otherwise $\exists n$ such that $a_n=l$ and $l=a_n\not\in A_n$. So if the limit point $l$ exists, it is outside of $A$. So we found an element $l\not\in A$, thus $A$ is a proper subset of $X$.

The problem I have here is to see if there indeed exists such limit point $l$ in $X$ or not. Since Rudin didn't cover sequences yet (it is chapter 2, basic topology), So I think I've constructed a Cauchy sequence but not sure it converges in $X$.

Please help me to finish this up.

user642721
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  • I know that this exercise is asked several times, and I read them all. But what I want to do is to finish my own (tedious) proof. – user642721 May 01 '19 at 00:39
  • I think your contrapositive is wrong. The statement is: If $X$ is a connected metric space with two points, then $X$ is uncountable. So the contrapositive is: If $X$ is countable (or possibly finite), then $X$ is not a connected metric space. In this sense, $X$ could be a metric space and fail to be connected or it could be connected and fail to be a metric space. – Clayton May 01 '19 at 00:44
  • @Clayton Not contrapositive of that statement, what I meant was contrapositive of "$X$ connected, $A$ countable, $A$ open $\implies$ $A$ not closed". Since "$X$ connected" and "$A$ countable" remain the same, I thounght that the contrapositive of that would be "$X$ connected, $A$ countable, $A$ closed $\implies$ $A$ not open". – user642721 May 01 '19 at 00:48
  • I'm sorry, I did not read carefully enough. You want to finish your proof. – Severin Schraven May 01 '19 at 00:56
  • The sequence you construct does not even need to be Cauchy. Consider $${ 2 \sum_{j=1}^n 1/j \ : \ n\in \mathbb{N} }.$$ – Severin Schraven May 01 '19 at 01:07
  • Even if you would force it to be Cauchy, it will not need to converge. Consider $$X= {1/(n+1) \ : \ n\in \mathbb{N} } $$ then the sequence $(1/(n+1))_{n\in \mathbb{N}}$ will not converge in $X$. You would need the additional assumption that $X$ is complete. – Severin Schraven May 01 '19 at 01:12
  • @SeverinSchraven But your $X$ is disconnected, isn't it? And why the above sequence is not Cauchy? I guess for any given $\epsilon>0$ we can find $N>0$ such that $k,i>N\implies 2\sum_{j=i}^k 1/j<\epsilon$...? – user642721 May 01 '19 at 01:18
  • @SeverinSchraven Oh maybe not. It was the harmonic series – user642721 May 01 '19 at 01:20
  • @SeverinSchraven How about to narrow down the radius of the ball from $1/i$ to $1/2^i$? Would it be sufficient? – user642721 May 01 '19 at 01:23
  • Narrowing down the radius in that way makes your sequence Cauchy, but that doesn't mean it converges. Consider $$ { 4^{-n} \ : \ n\in \mathbb{N}}$$ Then $(4^{-n})_{n\in \mathbb{N}}$ is Cauchy, but not convergent. – Severin Schraven May 01 '19 at 02:29
  • Also, what do you mean by "your $X$ is disconnected"? I proved in my answer that it must be disconnected, otherwise the statement in your question would be wrong. I'm only providing examples where your proof doesn't apply. – Severin Schraven May 01 '19 at 02:32
  • @SeverinSchraven Why $(4^{-n})$ is not convergent? Doesn't it converge to $0$? – user642721 May 01 '19 at 04:36
  • Also, your proof is an example of "$X$ disconnected $\implies$ there is a sequence that doesn't converge in $X$". What I wanted to show was "$X$ connected $\implies$ there is a sequence converging into $X$." I don't understand why this is a counterexample. – user642721 May 01 '19 at 04:43
  • First of all, no, the sequence I wrote does not converge to zero, because zero is not an element in the metric space I have chosen. Second, my example shows that your proof is simply wrong as stated. You gave an algorithm how to chose the sequence that should converge and I gave you an example where this is not the case. This doesn't mean the statement you want to prove is wrong, but your proof is. – Severin Schraven May 01 '19 at 09:36

1 Answers1

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Pick your favourite point $x_0$ and your second favourite point $x_1\neq x_0$. Assume your metric space $(X, d)$ is countable. Then the set $$S:=\{ d(x_0, y)\in \mathbb{R} \ : \ y\in X\} $$ is countable. This means that we can pick $r\in \mathbb{R}\setminus S$ such that $0<r< d(x_0, x_1)$. However, now $$ M:=B(x_0, r)= \overline{B(x_0, r)}$$ as there are no points with distance $r$ from $x_0$. Thus, $M$ is a nonempty clopen set with nonempty complement and therefore $(X,d)$ is not connected.

Severin Schraven
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