### CSIR DEC 2015 PART B SOLUTION

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:

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

$2. A''=A'.$

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

$4.A'=X.$

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.

### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...