For a subset $A$ of the topological space $X$, let $A'$ denotes the union of set $A$ and all those connected components of $X-A $ which are relatively compact in $X$ (i.e. the closure is compact in $X$). Then for every subset $A$ of $X$, what can we say:

$1. A' \text{ is compact.}$

$2. A''=A'.$

$3. A' \text{ is connected.}$

$4.A'=X.$

3. (

4.

2. (

$1. A' \text{ is compact.}$

$2. A''=A'.$

$3. A' \text{ is connected.}$

$4.A'=X.$

**Solution :****1. (**

**false**) Let $X = (0,1)$ (**any non-compact set will work**) with usual topology and let $A = X$ then $X - A = \phi$ empty set. Therefore $A^{'} = X$ which is not compact.3. (

**false**) Let $X = (0,1) \cup \{2\}$ and $A = (0,1)$ then $\{2\}$ is both open and closed in $X$. Now $X - A = \{2\}$ and so $A^{'} = X$ but X is not connected.4.

**(false)**Let $X = (0,1) \cup (1,2)$ and $A = (1,2)$ then $X-A = (0,1)$ is not complact in X. Therefore $A^{'} = A \ne X$.2. (

**True**) Its a CSIR part B problem, so one option should be true. Try the proof by yourself. It is immediate from the definition. Any doubts, happy to help you in the comments section.**Share to your groups:****(Any doubt in the argument comment below)**
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