Neighborhood of each number

**dwsmith** · June 16th 2011, 09:34 AM

Suppose x is real and epsilon > 0. Prove that $(x-\epsilon, \ x+\epsilon)$ is a neighborhood of each of its numbers; in other words, if $y\in (x-\epsilon, \ x+\epsilon)$ , then there is a delta > 0 such that $(y-\delta, \ y+\delta)\in (x-\epsilon, \ x+\epsilon)$ .

So since $y\in (x-\epsilon, \ x+\epsilon)$ , then $x-\epsilon < y< \ x+\epsilon\Rightarrow |y-x|<\epsilon$

I am not sure what to do after that.

**Moo** · June 16th 2011, 09:48 AM

Hi,

For simplification, let's say that y belongs to (x-e;x). So y>x-e. By letting $\delta=\frac{y-(x-\epsilon)}{2}$ , we find a $\delta$ such that $(y-\delta,y+\delta) \subset (x-\epsilon,x+\epsilon)$ (note that it's a subset, not "belongs to")

For the general stuff, you may just consider $\delta=\frac{\min (|y-(x-\epsilon)|,|(x+\epsilon)-y|)}{2}$ and it should work (I hesitated between min and max, but I think it's definitely min).

**Plato** · June 16th 2011, 09:48 AM

Originally Posted by dwsmith

Suppose x is real and epsilon > 0. Prove that $(x-\epsilon, \ x+\epsilon)$ is a neighborhood of each of its numbers

If $y\in(x-\epsilon, \ x+\epsilon)$ and $y\ne x$ then let $\delta=\frac{\min\{|x-y|,\epsilon -|x-y|\}}{2}$ .

**Deveno** · June 17th 2011, 04:58 AM

Originally Posted by Plato

If $y\in(x-\epsilon, \ x+\epsilon)$ and $y\ne x$ then let $\delta=\frac{\min\{|x-y|,\epsilon -|x-y|\}}{2}$ .

i would think that a formula like:

$\delta = {\min\{y-x+\epsilon,x-y+\epsilon\}$

would be preferable, as it places no restrictions on y.

note that both of these numbers have to be positive, since

|y-x| < ε. also, one of these numbers will be less than ε, unless y = x,

in which case choosing δ = ε works just fine.

noting the extreme case where y = x+ε - ε' (where ε' << ε),

x-y+ε = x-x-ε+ε'+ε = ε', whereas

y-x+ε = x+ε-ε'-x+ε = 2ε-ε' > ε, so this is clearly not the minimum.

the other "extreme" case is similar:

y = x-ε + ε' (where ε' << ε), and we have

y-x+ε = x-ε+ε'-x+ε = ε', but

x-y+ε = x-x+ε-ε'+ε = 2ε-ε' > ε, again, clearly the larger number of the two.

i see no reason to "divide by 2", the total length of the interval (x-ε,x+ε) is already 2ε.

**Tinyboss** · June 17th 2011, 05:20 AM

If you're allowed to use topological arguments, just say that $U=(x-\varepsilon,x+\varepsilon)$ is a basis element of the metric topology, and so for every point inside it, there's another basis element centered at that point and contained in U. (Using the standard basis of $\varepsilon$ -balls at each point.)

**Deveno** · June 17th 2011, 06:13 AM

i suspect that this question is part of an equivalence: U is open iff it is a neighborhood of each of its points.

(in this particular question, we are proving an open interval is open. it seems obvious, but it's worth proving at least once in one's life).

depending how "open" and "neighborhood" are defined, this can range from difficult, to axiomatic (a tautology).

**Plato** · June 17th 2011, 06:21 AM

Originally Posted by dwsmith

Suppose x is real and epsilon > 0. Prove that $(x-\epsilon, \ x+\epsilon)$ is a neighborhood of each of its numbers.

Originally Posted by Deveno

i would think that a formula like:
$\delta = {\min\{y-x+\epsilon,x-y+\epsilon\}$
would be preferable, as it places no restrictions on y.

Well yes, but I was building off of reply #2.
What would be preferable is $\delta=\epsilon-|x-y|>0.$
There are no restrictions on y.

If $t\in(y-\delta, \delta+y)$ then
$|x-t|\le |x-y|+ |y-t|<|x-y|+\delta=\epsilon.$

So $(x- \epsilon , \epsilon +x)$ is a neighborhood of each of its numbers.

**Deveno** · June 17th 2011, 06:39 AM

yes, your formulation of δ is a bit more elegant (although equivalent) than mine (because it takes fewer characters to write).

although i think you meant if: $t \in (y-\delta,y+\delta)$ , since your interval may not even be a subinterval of the ε-ball centered at x (for example, if x is very large).

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