IIT Jam 2019 Series Question

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   Solution is given(explained) in following YouTube video


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Csir net june 2018: Real analysis Question (1)

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General Mathematics Question : Given $(x_n)$ is convergent sequence in $\mathbb{R}$ and $(y_n)$ is bounded sequence in $\mathbb{R}$ then we can conclude that, (a) $(x_n+y_n)$ is convergent. (b) $(x_n+y_n)$ is bounded. (c) $(x_n+y_n)$ has no convergent subsequence. (d)$(x_n+y_n)$ has no bounded subsequence. Solution : we discuss options one by one (a) take $x_n=1$ for all $n\in\mathbb{N}$ and $y_n=(-1)^n$ for all $n\in\mathbb{N}$ then, sequence $(x_n)$ is convergent and $(y_n)$ is bounded. But, $(x_n+y_n)=(0,2,0,2,0,2,....)$ which is not convergent. Hence (a) is false. (b) given that sequence $(x_n)$ is convergent and hence it is bounded. Also given that, $(y_n)$ is bounded. So that sum of two bounded sequence $(x_n+y_n)$ is also bounded. Hence (b) is true. (c), (d): By option (b) as sequence $(x_n+y_n)$ is bounded sequence and hence by "Bolzano Weierstrass theorem for sequences", it has convergent (hence bounded) subsequence.  So that, (c),(d) are fa...
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IIT JAM 2018 Real analysis Question (2)

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General Mathematics Question Solution: we discuss options one by one: (A) if $P$, and $Q$ are compact subsets of $\mathbb{R}$ then $P$ and $Q$ both are bounded and closed subsets of $\mathbb{R}$ (By "Heine-Borel theorem"). Also we know that,  Union of two bounded subsets of $\mathbb{R}$ is bounded subset of $\mathbb{R}$ Finite union of closed sets in $\mathbb{R}$ is again closed set in $\mathbb{R}$. Hence, $P\cup Q$ is again bounded and closed subset of $\mathbb{R}$ and hence by "Heine Borel theorem" it is compact. So that option (A) is true. (B) take $P=\mathbb{Q}$ and $\mathbb{Q^c}$ then both $P$ and $Q$ are nonempty disjoint subsets of $\mathbb{R}$ that are not connected, but their union is $\mathbb{R}=(-∞,+∞)$ is connected subset of $\mathbb{R}$. Hence (B) is false.  (C) take $P=\{1,2\}$, $Q=(1,2)$ then $P$ is closed, and $P\cup Q=[1,2]$ is closed. Also both $P,Q$ are nonempty disjoint subsets of $\mathbb{R}$. But, $Q$ is not closed subset of $\m...
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