# If X has a countable basis then uncountably many points of A are limit points of A

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!

Question:

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.

Proof:

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.

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

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.