Need help formalizing "T is an open set"

In summary, to show that T is an open set, you need to find an interval around P where every point is in T.
  • #1
317
26

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.

Homework Equations


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 :cry:
 
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  • #2
Terrell said:

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.

Homework Equations


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 :cry:
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.
 
  • #3
Mark44 said:
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?
 

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