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This is an exercise in Munkres's book of topology.

If $X$ is a connected metric space and there are at least two points in $X$, then $X$ is not countable.

I have attempted to find the proof by constructing a chain of non-empty closed sets, say $K(i+1)$ belongs to $K(i)$, and $x_i$ lies out side $K_i$. But I failed to show that the intersection of all the $K_i$'s should not be empty because $X$ is not necessarily compact.

Seirios
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Li Xinghe
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  • Oh, I solved the question just now. Suppose a and b are the two points. Define the set D(x) to be {c|d(c,a)<x}. Then for every x in (0, d(a,b)), the set D(x) must be non-empty. If not, then D(x) and {y|d(a,y)>x} will be two disjoint open sets whose union is the whole space. – Li Xinghe Oct 31 '13 at 07:17

1 Answers1

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Pick a point $x_1$ in your space and cnsider the function $f:x\in X\mapsto d(x,x_1)\in\mathbb R$.

Since $X$ has two points, this function takes at least two values and since $X$ is connected, the image of $f$ is connected. It follows that the image of $f$ contains, at least, a non-empty closed interval.

Conclude now what you want.

Mariano Suárez-Álvarez
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