$A\subset \mathbb{R}$ with more than one element and $A/ \{a\}$ is compact for a fixed $a\in A$

Question

Question is :

Suppose $A\subset \mathbb{R}$ with more than one element and $A/ \{a\}$ is compact for a fixed $a\in A$ then

$A$ is compact
Every subset of $A$ must be compact
$A$ must be finite
$A$ is disconnected

Only compact subsets of $\mathbb{R}$ I can think of are finite union of closed intervals and finite sets.

Take a finite union of closed intervals, If i remove one element for that I would end up with something which contains $[a,b)$ and this would be not compact so should be the whole set.

So, First option is wrong i.e., $A$ is not compact.

I can not say anything about second and fourth option but third option is possible I guess.

Please help me to see this in detail..

No, your property is false, consider $A=[0,1]\cup{2}$ and $a=2$ — Camilo Arosemena-Serrato, Dec 22 '13 at 13:49
@Praphulla Koushik : the first bullet point is true because the union of two compact sets is compact. — Stefan Smith, Dec 22 '13 at 14:21

score 4 · Accepted Answer · answered Dec 22 '13 at 14:04

4

Let $\{U_i\}_{i\in I}$ be an open cover for $A$. This is also an open cover for $A\setminus\{a\}$. As this is compact, there exists a finite subcover. Together with one $U_i$ with $a\in U_i$ we obtain a finite subcover for $A$. Hence, yes, $A$ is compact.
No. Consider $A=[0,1]\cup\{2\}$, which has the noncompact subset $(0,1)$.
No. Consider $A=[0,1]\cup\{2\}$ again, which is infinite.
Yes. Consider the open cover $$A\setminus\{a\}\subseteq\bigcup_{r>0}(\mathbb R\setminus[a-r,a+r]).$$ By compactness there is a finite subcover, which boils down to $A\setminus\{a\}\subseteq \mathbb R\setminus[a-r,a+r]$ for some $r>0$. Thus $a$ is an isoloated point of $A$ and $A$ is disconnected.

answered Dec 22 '13 at 14:04

Hagen von Eitzen

374,180

Perfect.... Thank you :) – Dec 22 '13 at 14:48
Another way to see that $a$ is isolated in $A$: if not, there would be a sequence $(x_n)$ from $A\setminus{a}$ that converged to $a$. This sequence would contradict sequential compactness of $A\setminus {a}$. – Henno Brandsma Dec 28 '13 at 15:31
How $a$ being isolated point implies $A$ disconnected? Can you explain a bit more to me? – Kushal Bhuyan Oct 15 '15 at 13:24

score 3 · Answer 2 · answered Dec 22 '13 at 13:53

3

Hints:

The set $\;[0,1]\cup\{2\}\;$ already deals with options 2,3,4...

Option 1 is true, otherwise there exists $\;\{a_n\}\subset A\;$ s.t. that no subsequence of it converges in $\;A\;$, so...

answered Dec 22 '13 at 13:53

DonAntonio

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Yes Yes... I got it :) Thank you :) – Dec 22 '13 at 14:49

$A\subset \mathbb{R}$ with more than one element and $A/ \{a\}$ is compact for a fixed $a\in A$

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