Need help formalizing "T is an open set"

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SUMMARY

The discussion focuses on proving that the set T, defined as T = { t ∈ ℝ : ∃ s ∈ S, |t - s| < ε}, is an open set in the context of real analysis. The user attempts to demonstrate this by showing that for any point x in T, there exists a δ such that the interval (x - δ, x + δ) is contained within T. The key to the proof lies in establishing the relationship between ε and δ, ensuring that |σ - y| < ε holds for all y in the interval (x - δ, x + δ). The discussion highlights the importance of correctly applying the triangle inequality to simplify the expressions involved.

PREREQUISITES
  • Understanding of open sets in topology
  • Familiarity with the ε-δ definition of limits
  • Knowledge of the triangle inequality in real analysis
  • Basic concepts of intervals in the real number line
NEXT STEPS
  • Study the properties of open sets in metric spaces
  • Learn about the ε-δ definition of continuity and limits
  • Explore the triangle inequality and its applications in proofs
  • Investigate the differences between open and closed intervals in real analysis
USEFUL FOR

Students of real analysis, mathematicians focusing on topology, and anyone interested in understanding the properties of open sets and their proofs.

Terrell
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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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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.
 
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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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