General topology, compact sets, neighborhoods

Question

Consider the following statements.If a statement is true supply a proof; if false give a counterexample. (a) Two distinct points p and q can be separated by disjoint neighborhoods. (b) A set K = {p1, p2, ..., pk} and a point q that doesn't belong to K, can be separated by two disjoint open sets. (c) A set K = {p1, p2, ...} and a point q that doesn't belong to K, can be separated by two disjoint open sets

I have to solve these. I think that the first two are true and the last one is false. But I need powerful proofs for each of them.

these are the proofs that i need to strengthen and use more details: 1) If p≠q then d(p,q)>0 and for ϵ=12d(p,q)>0 the sets Up={x∈X∣d(p,x)<ϵ} and Uq={x∈X∣d(q,x)<ϵ} are disjoint open sets containing p and q respectively.

2) Similarly for i=1,…,k let Ui denote an open set containing pi and let Vi denote an open set containing q with Ui∩Vi=∅. Now take the union of the ⋃ni=1Ui and V=⋂ni=1Vi.

3)Counterexample: R with usual topology with pi=1i and q=0.

Answer 1

For b)Let $M=\bigcup_{i=1}^{k}U_\epsilon(p_i)$ where $U_\epsilon(p_i)$ is the open ball around $p_i$. By definition $M$ is the union of open sets therefore open. Denote $N=U_\epsilon(q)$.

You want to show that $N\cap M =U_\epsilon(q)\cap \bigcup_{i=1}^{k}U_\epsilon(p_i)=\emptyset$ which is nothing less than checking if $U_\epsilon(q)\cap U_\epsilon(p_i)=\emptyset$ for every $i=1,...,k$. Which leads to $\epsilon=...$

For c) I do not know how to understand your example, but I would take $K=(0,q)\cup(q,1)$ meaning all elements in between $0$ and $1$ except for $q$. There is no possibility to put $q$ into an open set which would be disjoint to K.

Answer 2

Let me restate your question to be sure I understand it. All statements assume that the underlying topology is a metric space.

(a) Two distinct points $p$ and $q$ can be separated by open sets.

(b) A finite set $K$ and a point $q$ that's not a member of $K$ can be separated by open sets.

(c) A countably infinite set $K$ and a point $q$ that's not a member of $K$ can be separated by open sets.

If I'm correctly understanding your question, then your intuition is correct. (a) and (b) are true and (c) can be false.

(a) Let $a = d(p,q)$. Then the open sets $B(p, a/4)$ and $B(q,a/4)$ separate $p$ and $q$.

(b) Let $a = min\{d(p_i, q)\mid i = 1, 2, \ldots k \}$. This minimum exists and is positive because $K$ is finite. Then for each $i$, the open sets $U_i = B(p_i, a/4)$ and $B(q,a/4)$ separate $p_i$ and $q$. Let $U = \cup \{U_i \mid i = 1, 2, \ldots k \}$. Then $U$ is a union of open sets, so it is itself open, and $U$ and $B(q,a/4)$ separate $K$ and $q$.

(c) In $\Bbb R$, let $p_n = 1/n$ and let $q = 0$. Then any open set $U$ containing $q$ contains $B(q, \epsilon)$ for some $\epsilon \gt 0$. Choose $n$ with $1/n \lt \epsilon$. Then $p_n \in U$ so $U$ does not separate $q$ and $K$.