Please note that the title is not the complete question I am asking, it just serves as a topic summary. The question is below.

I am hoping someone can review my proof. I have put a (*) next to the claim I am personally most uncertain about. Thanks in advance!


Let X have a countable basis; let A be an uncountable subset of X. Show that uncountably many points of A are limit points of A.


Let X be a topological space with a countable basis. Let A be an uncountable subset of X.

Let A' be the set of limit points of A that are contained within A.

Suppose A' is not uncountable. Then A' is countable.

Since A is uncountable, we have B = A - A' is non-empty.

(*) Also we have that B must be uncountable since if B was countable we would have:

$$ A = B \bigcup A' $$

But then A would be the union of two countable sets, which is countable. Yet A was chosen to be uncountable. So B must be uncountable.

So we have B $\subset$ A and B consists of an uncountable number of points each of which is a member of A yet not a limit point of A. Then we for each x$_i$ $\in$ B we can find an open set U$_i$ such that x $\in$ U$_i$ and U$_i$ $\bigcap$ A = {x$_i$}. Then since U$_i$ is open it must contain some basis element B$_i$ such that x$_i$ $\in$ B$_i$ $\subseteq$ U$_i$.

Now the set of B$_i$ basis elements is an uncountable set of basis elements of X since for each unique $x_i$ we produce a unique B$_i$. Consider that for some x$_n$ and x$_z$ that their corresponding open basis elements B$_n$ and B$_z$ cannot be the same basis elements since if either were, the basis element would contain both x$_n$ and x$_z$ which is not allowed since the points of B are limit points and the open set B$_i$ must only intersect A in one single point, that is x$_i$.

Hence we have an uncountable collection of basis elements in a space that has a countable basis. A contradiction. So A must contain uncountably many points that are limit points of A.

  • 2
    $\begingroup$ yes, it's correct. $\endgroup$ Mar 25, 2019 at 4:54
  • $\begingroup$ I think you meant to say "since the points of $B$ are not limit points. $\endgroup$
    – Math1000
    Nov 18, 2019 at 19:40

1 Answer 1


I concur with the comment that the proof is correct. You don't really need indices $x_i$: just index using the points: for each $x \in B$ find $U_x \in \mathcal{B}$ such that $U_x \cap A= \{x\}$ etc.

Maybe a slight rephrasing: $A$ as a subspace also has a countable base $\mathcal{B}$ (the intersections of the members of the countable base for $X$ with $A$) and $A = A_1 \cup A_2$, where $A_1 = \{x \in A: x \in A'\}$, the limit points in $A$ and $A_2 = \{x: x \text{ isolated in } A\}$ and the union is disjoint. Note that $x \in A_2$ implies $\{x\}$ open in $A$ which implies $\{x\} \in \mathcal{B}$. So $A_2$ embeds injectively into $\mathcal{B}$ via $x \to \{x\}$, so $A_2$ is countable. It follows that $A_1$ is uncountable as otherwise $A$ would be countable as a union of two countable sets. QED.


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