# Need help formalizing "T is an open set"

## Homework Statement

Let $S\subseteq \Bbb{R}$ and $T = \{ t\in \Bbb{R} : \exists s\in S, \vert t-s\vert \lt \epsilon\}$ where $\epsilon$ is fixed. I need to show T is an open set.

n/a

## The Attempt at a Solution

Let $x \in T$, then $\exists \sigma \in S$ such that $x \in (\sigma -\epsilon, \sigma +\epsilon)$. Let $\delta = \min(\frac{\vert \sigma -\epsilon -x\vert}{2}),\frac{\vert \sigma +\epsilon -x\vert}{2}$ and consider the interval $(x-\delta, x+\delta)$. If $y \in (x-\delta, x+\delta)$, then
\begin{align}
\vert \sigma - y \vert \leq \vert \sigma -x\vert + \vert x- y\vert \lt \epsilon + \delta \cdots
\end{align}
I want to show $\vert \sigma -y \vert \lt \epsilon$, but I am having trouble with finding the right expression to simplify this

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Mark44
Mentor

## Homework Statement

Let $S\subseteq \Bbb{R}$ and $T = \{ t\in \Bbb{R} : \exists s\in S, \vert t-s\vert \lt \epsilon\}$ where $\epsilon$ is fixed. I need to show T is an open set.

n/a

## The Attempt at a Solution

Let $x \in T$, then $\exists \sigma \in S$ such that $x \in (\sigma -\epsilon, \sigma +\epsilon)$. Let $\delta = \min(\frac{\vert \sigma -\epsilon -x\vert}{2}),\frac{\vert \sigma +\epsilon -x\vert}{2}$ and consider the interval $(x-\delta, x+\delta)$. If $y \in (x-\delta, x+\delta)$, then
\begin{align}
\vert \sigma - y \vert \leq \vert \sigma -x\vert + \vert x- y\vert \lt \epsilon + \delta \cdots
\end{align}
I want to show $\vert \sigma -y \vert \lt \epsilon$, but I am having trouble with finding the right expression to simplify this
IMO you're using too many symbols: x, y, $\sigma, \epsilon, \delta$. Let's say you have an interval T of radius r, centered at P: (P - r, P + r). For any point x in this interval, there is some $\epsilon$ so that every point in the interval $(x - \epsilon, x + \epsilon)$ is also in the interval T. The same would not be true of closed intervals.

For any point x in this interval, there is some ϵϵ\epsilon so that every point in the interval (x−ϵ,x+ϵ)(x−ϵ,x+ϵ)(x - \epsilon, x + \epsilon) is also in the interval T.
Am I trying to do too much by using the triangle inequality?