Compactness and connectedness of sets

Question

Let $$S=\{(x,y)\in \mathbb{R}^{2} \mid -1\leq x\leq 1,\ -1\leq y\leq 1 \}$$ and let $$T=S\setminus\{(0,0)\},$$ then what about the compactness and connectedness of $S$ and $T$?

Since $S$ is closed and bounded subset of $\mathbb{R}^{2}$, by Hein Borel theorem it is compact. Since there does not exist a separation by open sets $S$ is connected also. And $T$ is obtained by removing the point $(0,0)$. Then how it effects the compactness and connectedness of $T$? Is the same effect when removing any point?

Answer 1

Compactness : In general, removing a point does not preserve compactness.

See that $[0,1]$ is compact but $(0,1]$ is not compact.

You can consider sequence $\left(\frac{1}{n}\right)$ that converges but not to a point in $(0,1]$.

Similarly, the square $S$ you have mentioned is compact, but, resulting space obtained by removing the point $(0,0)$ is not compact. You can imitate the example that I have given to produce a sequence that converges but not in $T$.

Connectedness : Removing $1$ from interval $[0,2]$ breaks the connectedness of $[0,2]$. This can be seen clerarly from drawing that line and removing that point.

Removing line $\{(x,x):x\in \mathbb{R}\}$ from $\mathbb{R}^2$ is not connected as you can see visually.

In general, removing what is called n dimensional subset from a $n+1$ dimensional connected space usually leaves it disconnected.

But, removing a point ($0$ dimensional) from $S$ ($2$ dimensional space) keeps it connected.

Answer 2

(1). Compactnes: In a $T_2$ (Hausdorff) space (such as $\Bbb R^2$) a compact subset must be closed. But $T$ is not closed. Also $S\setminus \{p\}$ is not closed for any $p\in S,$ because $p\in \overline {S\setminus \{p\}}$.

Let $T$ be a non-closed subset of a Hausdorff space $X$ and let $p\in \overline T \setminus T.$ To prove that $T$ is not compact:

For each $t\in T$ let $U_t$ and $V_t$ be disjoint open sets with $t\in U_t$ and $p\in V_t.$ Then $C=\{U_t: t\in T\}$ is an open cover of $T.$ If $I$ is a finite subset of $T$ and $D=\{U_t: t\in I\}$ then:

(i). If $I=\emptyset$ then $D$ is not a cover of $U$ because $U$ is not empty because $U$ is not closed.

(ii). If $I \ne \emptyset$ then $W=\cap \{V_t:t\in I\}$ is an open set containing $p.$ So $W\cap T\ne \emptyset$ (because $p\in \overline T$), but $W$ is disjoint from $\cup D.$ So $D$ is not a cover of $T.$

(2). Connectednes: Suppose $A,B$ are open subsets of $\Bbb R^2$ such that $A\cup B\supset T=[-1,1]^2\setminus \{(0,0)\}$ and $(A\cap T)\cap (B\cap T)=\emptyset. $

For $r\in (0,1)$ let $C(r)=T\setminus (-1,r)^2$ and $C^-(r)=T\setminus (-r,1)^2.$

For brevity let $D(r)=C^+(r)\cup C^-(r)=T\setminus (-r,r)^2.$

Each of $C^+(r),\;C^-(r)$ is connected so each is a subset of $A$ or a subset of $B.$ And $C^+(r)\cap C^-(r)\ne \emptyset$ , so $D(r)$ is a subset of $A$ or a subset of $B. $

For any $r,r'\in (0,1)$ we cannot have $D(r)\subset A$ and $D(r')\subset B,$ because $D(r)\cap D(r') \ne \emptyset .$ So either $D(r)\subset A$ for all $r\in (0,1)$ or $D(r)\subset B$ for all $r\in (0,1).$

Now $T=\cup_{r\in (0,1)} D(r)$ so $T\subset A$ or $T\subset B.$

A similar argument works for $S\setminus \{p\}$ for any $p\in (-1,1)^2,$ and with a little modification, for any $p\in S.$ This method will also work in higher dimensions.

Remarks: If $P,Q$ are connected subsets of a space $X$ and $P\cap Q\ne \emptyset$ then $P\cup Q$ is connected. So $C^(r)=([-1,1]\times [r,1])\cup ([r,1]\times [-1,1])$ is connected, and similarly $C^-(r)$ is connected.

Also if $F$ is a family of connected sets such that $a\cap b\ne \emptyset$ whenever $a,b\in F$, then $\cup F$ is connected. We can apply this with $F=\{D(r):r\in (0,1)\}.$

Answer 3

$ S$ is compact due to Heine-Borel theorem, as you mentioned, and it is also connected. You can prove this in several ways. One way is considering that $ S$ is the topological product of two connected spaces, another is to show that $S$ is path-connected (hint: it's also convex).

$T$ is not compact, because Heine-Borel theorem states that a subspace of $ \mathbb{R}^n $ is compact if and only if it is closed and bounded. Now, $(0,0)$ is adherent to $T$, and it does not belong to $T$, therefore $T$ is not compact.

However, $T$ is connected, and even path-connected, but it's not convex, so we have to adapt our previous proof. Consider $A,B\in T$. If the straight line joining $A$ and $B$ doesn't meet $(0,0) $, there is no problem. If it does, you can change the path a little bit having it go on a small semicircumference centered in $(0,0) $. In general, subtracting one point does not preserve connectedness.