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

  1. $A$ is compact
  2. Every subset of $A$ must be compact
  3. $A$ must be finite set
  4. $A$ is disconnected

for any open cover of $A$ is also open cover for $A-\{a\}$, since $A-\{a\}$ is compact implies that it has finite sub cover to cover $A-\{a\}$. with this sub cover union an open set containing $a$ will be the finite sub cover for $A$. Hence $A$ is compact

Given that $A-\{a\}$ is compact, so it is closed, this implies $\{a\}$ is open. But every singleton in $\mathbb{R}$ is closed, so $\{a\}$ is both closed and open. Hence A is disconnected

so option (1) and (4) is right. what can we say about option (2) and (3)?

vijayanand
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  • Hint: Let $B=A-{a}$. Then $A=B\cup {a}$ – JMoravitz Nov 24 '16 at 07:17
  • Surely there are no reasons for 2 and 3 to be true and you can come up with counter examples. Let A-{a} be ... any compact set ... say [0,1] and so A = [0,1] U {a}. Surely this has non compact subsets and surely is not finite. – fleablood Nov 24 '16 at 08:38
  • that $a$ removed from $A$ is not fixed one. $A-{a}$ is compact for any $a \in A$, so if i take a element from that closed interval, then the set remains not compact. a contradiction to the hypothesis. – vijayanand Nov 24 '16 at 09:06

1 Answers1

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The proof that 1) is true is ok.

But if $A=\{a\}$ then $A\setminus \{a\}$ is compact and $A$ is connected. So 4) is false.

Counterexamples to 2) and 3) are given in comments: $A=[0,1]\cup\{3\}$ and $a=3$

user126154
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