Prove that, for $x \in \mathbb R$ and $\delta_x > 0$, the open interval $(x-\delta_x, x+\delta_x)$ is itself an open set.

I am preparing for my exam and we will be asked to prove various theorems and I am struggling a lot with them. Frankly, I don’t have an idea of how to start. Can anyone please help with this proof?

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    $\begingroup$ When I took a first course in topology, this was the definition of an open set in $\mathbb R$, since $\mathbb R$ is a metric space. $\endgroup$ Jan 25, 2015 at 8:37
  • $\begingroup$ How have you had the topology on the reals defined? $\endgroup$ Jan 25, 2015 at 8:37

1 Answer 1


Let $y \in (x-\delta_x,x+\delta_x)$. We must show that $y$ is an interior point of $(x-\delta_x,x+\delta_x)$.

We have $|y - x| < \delta_x$. Therefore $r = \delta_x - |y-x| > 0$. We will prove that $(y-r,y+r) \subseteq (x-\delta_x,x+\delta_x)$, which will prove what we want.

Let $z \in (y-r,y+r)$. Then $$|z - x| \leq |z - y| + |y - x| < r + |y - x| = \delta_x.$$ This proves that $z \in (x-\delta_x,x+\delta_x)$, hence that $(y-r,y+r) \subseteq (x-\delta_x,x+\delta_x)$. Therefore $y$ is an interior point of $(x-\delta_x,x+\delta_x)$, completing the proof.


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