Proving that the set is open set of $R$ with respect to usual metric.

Question

Hi I had a midterm last week, but something is bothering me about my midterm questions so I decided to come here and ask.

1)Prove that the following set is open subset of $R$ with respect to the usual metric. $(-10, \infty)$

2) Prove that for every $x \in R$, the set $\{x\}$ is a closed in $R$ with respect to the usual metric.

My approach to this questions:

1. Union of open set is open too. So we need to show $(-10, \infty)$ as open sets. $(-10,-8)\cup(-9,-7)\cup(-8,-6)... $ But my assistant said that I need to show that set as a union of open balls. (Like $ \text{Bdu}(-9,1)\cup \text{Bdu}(-8,1)\cup...)$ So my answer is wrong.

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2. $R\backslash\{x\}=(-\infty,x)\cup(x,\infty)$

$(x,\infty)=(x,x+2)\cup(x+1,x+3)\cup(x+2,x+4)...$

$(-\infty,x)=(x-2,x)\cup(x-3,x-1)\cup(x-4,x-2)... $The set $R\backslash\{x\}$ is open, then $\{x\}$ is closed.

I got full marks from this question.

So I wonder, how can the second question be true if the first one is wrong? I am confused. Please excuse any mistakes, English is not my native language and I'm new here.

Answer 1

But my assistant said that I need to show that set as a union of open balls.

Well, this depends on how it was taught in your course. If you were taught that the basic open sets were $(a,b)$ for each $a<b$, then your answer is correct. If you were taught that they were $\mathrm{B}(x,r)$ for each $x$ and $r>0$, and you haven't shown that $\mathrm{B}(x,r)$ and $(x-r, x+r)$ are the same thing, then your answer is incorrect. We're unlikely to be able to advise without knowing the details of how you were taught.

how can the second question be true if the first one is wrong?

You're right that your approaches in both were the same. Perhaps your assistant didn't want to tell you off for the same mistake twice. Or perhaps the marks were allocated differently.

Answer 2

Assuming the first statement, the second follows by your proof. Your first answer looks fine to me although the iteration might not be very clear. An easier collection of balls might be $\cup_{n=1}^\infty (c,c+n)=(c,\infty)$, where $c\in R$, where $(c,c+n)=B_{n/2}(c+n/2)$, where $B_\delta(x):=\{y\in R: |x-y|<\delta \}$.