On compact sets Let $A$ be a subset of $\mathbb R$ with more than one element. Let $a\in A$. If $A\setminus \{a\}$ is compact, then


*

*$A$ is compact.

*every subset of $A$ must be compact.

*$A$ must be a finite set.

*$A$ is disconnected.


I think 1 is definitely correct. 2,3 may not be true. What about 4? Any hint please!
 A: Since we are in the reals, one can use the fact that a set of reals is compact if and only if it is closed and bounded. 
Suppose $A\setminus\{a\}$ is compact. Then it is bounded, so $A$ is bounded. And $A$ is the union of the two closed sets $A\setminus\{a\}$ and $\{a\}$.
The next two assertions are false. Let $A=[0,1] \cup \{17\}$, and let $a=17$. We leave finding a non-compact subset of $A$ to you.
For the last problem, we need to assume that $a\in A$. Consider the two open sets $X=(A\setminus\{a\})^c$ and $(\{a\})^c$, and use the definition of connected to conclude that $A$ is not connected. 
A: Working in $\mathbb{R}$ means that compact sets are closed and bounded. I am assuming that you mean $A\setminus \{a\}$ is compact for some $a$ as opposed to all $a$.
1) Yes, $A$ is compact. We know that $A\setminus \{a\}$ is compact, and hence closed. Since points in a Hausdorff space are closed, and the union of two closed sets is closed, we get $A = \{a\}\cup (A\setminus \{a\})$ is closed. It is also clear that $A$ is bounded, and hence compact.
2) This is not true. $[0,1]\cup \{2\}$ is compact but $(0,1)$ is not.
3) Again, this need not be true. $[0,1]\cup \{2\}$ is compact, and removing $\{2\}$ leaves a compact set. 
4) This is true for the same reason as 1). We can write $A = \{a\}\cup (A\setminus \{a\})$, a disjoint union of closed sets. In fact $\{a\}$ has to be an isolated point, otherwise removing it from $A$ would not leave something that is closed.
A: $\mathbb{R}$ is normal which means for every pair of disjoint closed sets $A, B $ there exists disjoint open sets $U$ and $V$ such that $A \subset U$ and $B \subset V$.  Since one point sets are closed in $\mathbb{R}$ and $A - \{a\}$ is closed because it's compact, you can find disjoint open sets $U, V$ that contain $A - \{a\}$ and {$a$} respectively.  Then $U\cap A$ and $V\cap A$ is a separation of $A$.
