I was just reading through a proof about the maximum possible cardinality of a separable Hausdorff space, but I'm stuck on one part of it. The essence of the proof is copied and pasted below, and they say "it's easy to see that $f$ is injective." I'm having trouble proving that $f$ is injective, probably because I'm new at topology.

Let $X$ be a separable Hausdorff space and let $A$ be a countable dense subset of $X$. Define a function $f:X\to\mathcal P(\mathcal P(A))$ by setting $f(x)=\{B\subseteq A:x\in\overline{B}\}$. Using the fact that $X$ is Hausdorff, it's easy to see that $f$ is injective, whence $|X|\le|\mathcal P(\mathcal P(A))|\le2^{2^{\aleph_0}}$.

My proof so far is as follows: We prove that $f$ is injective by contradiction. Suppose $f$ is not injective; that is, suppose $\exists x_1, x_2 \in X$ with $x_1 \neq x_2$ and $f(x_1) = f(x_2).$

$X$ is Hausdorff $\implies \exists U_1, U_2$ open in $X$ such that $x_1 \in U_1, x_2 \in U_2$, and $U_1 \cap U_2 = \varnothing$. Since $f(x_1) = f(x_2),$ then if $B \subseteq A$, $x_1 \in \overline B \iff x_2 \in \overline B$. We now consider 3 cases; note that cases 1 and 2 are not necessarily mutually exclusive.

Case 1: $\exists B\subseteq A: x_1 \in B$. Then consider $B_1 = \{x_1\} \subseteq A.$ $x_2 \notin B_1$ but $x_2 \in \overline{B_1}$, so $x_2$ is a limit point of $B_1$. But $U_2$ is an open set containing $x_2$ with $(U_2 \setminus \{x_2\}) \cap B_1 = \varnothing$, a contradiction.

Case 2: $\exists B \subseteq A: x_2 \in B$. Then consider $B_2 = \{x_2\} \subseteq A.$ $x_1 \notin B_2$ but $x_1 \in \overline{B_2}$, so $x_1$ is a limit point of $B_2$. But $U_1$ is an open set containing $x_1$ with $(U_1 \setminus \{x_1\}) \cap B_2 = \varnothing$, a contradiction.

Case 3: $\forall B \subseteq A$, $x_1 \notin B$ and $x_2 \notin B$. Let $B_0 \in f(x_1)$, then $x_1$ and $x_2$ must both be limit points of $B_0$....but now I'm stuck here, and not sure how to arrive at a contradiction. Please help!

  • 3
    $\begingroup$ For case 3, consider $B_0 \setminus U_1$ and $B_0\setminus U_2$. But you get it more directly, without any case distinction, by: Let $x_1 \neq x_2$. Then $\exists U_1,U_2$ ... Then consider $B_1 = A\cap U_1$. You will find that $B_1 \in f(x_1)\setminus f(x_2)$, so $f(x_1) \neq f(x_2)$. $\endgroup$ – Daniel Fischer Apr 16 '14 at 22:06
  • $\begingroup$ @DanielFischer, thank you very much! I've got it now! $\endgroup$ – justin Apr 16 '14 at 23:11
  • $\begingroup$ That's great. Did you do it both ways or only one? $\endgroup$ – Daniel Fischer Apr 16 '14 at 23:12
  • $\begingroup$ I did it with the more direct method you suggested; it seemed more elegant. $\endgroup$ – justin Apr 17 '14 at 2:07
  • $\begingroup$ @DanielFischer I typed up a solution below to illustrate how I used the direct method. If you are available to provide some more advice, there's a different topology problem I'm stuck on that I just posted here: math.stackexchange.com/q/757255/111520 $\endgroup$ – justin Apr 17 '14 at 3:16

Let $x_1, x_2 \in X$ with $x_1 \neq x_2$. $X$ is Hausdorff $\implies$ there exist open sets $U_1$ and $U_2$ such that $x_1 \in U_1, x_2 \in U_2$, and $U_1 \cap U_2 = \varnothing$. Define $B_1 \subseteq A$ as $B_1 = A \cap U_1$.

Then $x_2 \notin B_1$ since $x_2 \notin U_1$. Furthermore, $x_2$ is not a limit point of $B_1$, since $U_2$ is an open set containing $x_2$, and $(U_2 \setminus \{x_2\}) \cap B_1 = \varnothing$. Hence $x_2 \notin \overline{B_1}$.

Now we show that $x \in \overline{B_1}$. $x \in \overline{B_1} \iff \forall$ open sets $U$ containing $x$, $U\cap B_1 \neq \varnothing$. Then let $U$ be any open set containing $x$. $U\cap B_1 = U\cap (A\cap U_1) = (U\cap U_1)\cap A \neq \varnothing$, since $(U\cap U_1)$ is an open set containing $x$, and $A$ is dense in $X$.

$f(x_1) \neq f(x_2)$, so $f$ is injective.

  • $\begingroup$ One fact that would help shorten this exposition (and is perhaps implicit in what you have written above) is the following: if $D \subseteq X$ is dense and $U \subseteq X$ is open, then $\overline{ U \cap D } = \overline{U}$. (And this does not rely on any separation axiom holding in $X$.) $\endgroup$ – user642796 Apr 17 '14 at 9:16
  • $\begingroup$ Another thing that would help shorten the exposition is that $x\in \overline{B}$ if and only if every (open) neighbourhood of $x$ intersects $B$. Then the cases 1 and 2 are reduced to $U\cap B_1 = U\cap (U_1\cap A) = (U\cap U_1)\cap A \neq \varnothing$, since $U\cap U_1$ is an open set containing $x_1$, hence nonempty, and $A$ is dense. $\endgroup$ – Daniel Fischer Apr 17 '14 at 10:54
  • $\begingroup$ I see, thanks! I hadn't encountered those facts yet. $\endgroup$ – justin Apr 17 '14 at 13:44

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