Proving $\forall x \in \Bbb R$ and $ \epsilon \gt 0$, $N(x;\epsilon) \backslash \{x\}$ is open.

Question

Prove: $\forall x \in \Bbb R$ and $ \epsilon \gt 0$, $N(x;\epsilon) \backslash \{x\}$ is open.

Proof:

Let $S = N(x;\epsilon) \backslash \{x\}$.

We want to show that $S = intS$.

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Let $y \in S$. If we take $\delta = \frac{\epsilon - |x-y|}{2}$, then $N(y;\delta) \subseteq S$.

So $y \in intS$.

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Let $ y\in intS$. Then there exists an epsilon-neighborhood of $S$ such that $N(y;\epsilon) \subseteq S$.

So $y \in N(y;\epsilon) \subseteq S$ implies that $y \in S$.

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I want to know if my proof is correct.

Answer 1

All open balls $N(x; r), r>0$ are open in the metric topology, by definition. In any metric space $X$, $\{x\}$ is a closed set for any $x \in X$ and thus $\mathbb{R}\setminus \{x\}$ is open in $\mathbb{R}$.

So $N(x;r)\setminus \{x\} = N(x; r) \cap (\mathbb{R}\setminus \{x\})$ is open as the intersection of two open sets.

If you want a direct, more laborious, proof: for each $y \in N(x;r)\setminus\{x\}$, set $\delta=\min(d(x,y), r-d(x,y)) > 0$ and show that $y \in N(y; \delta) \subseteq N(x; r) \setminus \{x\}$.

Answer 2

The first part is actually sufficient as $\text{int}(S) \subseteq S$ always holds by definition. You made a small mistake when you defined $\delta$, but otherwise this is correct.

Answer 3

You can simplify your proof by noting that $\operatorname{int} X\subseteq X$ for any set $X$ in any topological space. By definition $\operatorname{int} X$ is the union of all open sets contained in $X$, so it too must be contained in $X$. Equivalently it is the set of all interior points of $X$, but all these elements are by definition in $X$, so once again the interior of $X$ must be contained in $X$.

For the other direction your $\delta$ has to depend on $y$ somehow. What if $y$ is in $N(x,\varepsilon/2)$? I'd suggest looking at $d(x,y)$.

Answer 4

I think that the $\delta$ might be wrong. There are some values for which $N(y, \delta)$ is not contained in $S$. Take $y = x + \frac{9}{10}\epsilon$ for example.