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Exercise: Using the fact $d$ is a metric hence continuous. Prove that if a metrizable space $(X,tau)$ is connected and $X$ has at least 2 points, then $X$ has uncountable number of points.

I found the following answer on Mathstackexchange by Mariano Suárez-Álvarez

Let your points be $a$ and $b$.

Let $lambdain(0,1)$ Suppose there is no point $x$ in your space such that $d(a,x)=lambda d(a,b)$. Then the sets ${z:d(a,z)<lambda d(a,b)}$ and ${z:d(a,z)>lambda d(a,b)}$ are two non-empty open sets which partition the space.

Since we are assuming connectedness, this is impossible.

Therefore, the image of the function $d(a,mathordcdot):Xtomathbb R$ contains the interval $(0,d(a,b))$, and this can only happen if $X$ is uncountable.

Questions:

1) Why does the author uses $lambda$? If instead the author used $d(a,b)$ and all other points were diferent from $b$ the distance would still be different, allowing the disjoint partition of the space.

2)
I know the metric is continuous function into $mathbb{R}$. How does the author deduces that the image contains $(0,d(a,b))$? How does that relate to the previous argument of the partition?

Thanks in advance!

I think your confusion arises because the proof you quote only uses continuity of the metric in a rather artificial way. Here are two proofs, the first doesn't really use the continuity and the second does. In both proofs, we assume $X$ is connected and metrizable and has at least two points, say $a neq b$.

1. Let $lambda in (0, 1)$. I claim that $d(a, z) = lambda d(a, b)$ for some $z = z(lambda)in X$, for, if not, the sets ${z:d(a,z)<lambda d(a,b)}$ and ${z:d(a,z)>lambda d(a,b)}$ are non-empty, open, disjoint and partition $X$, leading to a contradiction because $X$ is connected. Since there are uncountably many $lambda in (0, 1)$, and $d(a, b) neq 0$ there must be uncountable many $z(lambda) in X$. (This is the proof you quote adapted to remove the artificial appeal to continuity of the metric.).


2. Consider the function $f : X to Bbb{R}$ defined by $f(x) = d(a, x)$. $f$ is continuous and hence the image $I = f[X]$ is a connected subspace of $Bbb{R}$, because $X$ is connected and the continuous image of a connected set is connected. A non-empty connected subset of $Bbb{R}$ is either a single point or an interval. As $f(a) = d(a, a) = 0$ and $f(b) = d(a, b) neq 0$, $I$ is an interval and so contains uncountably many points. Thus the uncountable set $I$ is an image of $X$ and hence $X$ must be uncountable.

If it helps, the idea of the argument is two parts.

1) If $f: X to Y$ is a continuous map and $X$ is connected, then so is $f(X)$. It suffices to assume $f$ is surjective (otherwise, replace $Y$ with $f(X)$ with the subspace topology).

If $f$ is surjective, then given any pair of open sets $U_1, U_2 subset Y$ with $U_1 cap U_2 = varnothing$ and $U_1 cup U_2 = Y$, we may take the inverse image to obtain $f^{-1}(U_1)$ and $f^{-1}(U_2)$. These are nonempty if and only of the corresponding $U_i$ is nonempty; furthermore their intersection is trivial and union is all of $X$ (this does not use surjectivity). Therefore, if $U_1 cup U_2 = Y$ is a disconnection of $Y$ (that is, neither $U_i$ is the empty set), then the same is true of $f^{-1}(U_1) cup f^{-1}(U_2) = X$, and so if $Y$ is disconnected so is $X$, as desired.

2) If $I$ is a connected subset of the real line, if $a, b in I$, then $[a, b] subset I$. For if not, choose $x in (a,b)$ not in $I$, and write $I = left((-infty, x) cap Iright) cup left(x, infty) cap Iright)$ to obtain a disconnection of $I$.

Now consider the conditions map $X to [0,infty)$ given by $x mapsto d(a,x)$. This has image containing both $d(a,a) = 0$ and $d(a,b) neq 0$. So therefore the image must contain all of $[0, d(a,b)]$ and hence must be uncountable; so the domain is as well.

Mariano rolled these two points into one proof: he took the open sets $U_1$ and $U_2$ in the first step to be $(0, x)$ and $(x, d(a,b))$ if $x$ is not in the image.