Mathematics Asked on November 2, 2021
From Tao, Analysis II.
Consider the set $X := [1,2] cup [3,4]$, with the usual metric. This set is disconnected because the sets $[1,2]$ and $[3,4]$ are open relative to $X$.
This question starts from the same book section but I need more clarification. I will try to explain my guess of what’s going on.
For example, $1.5$ is an interior point of $[1,2]$ because letting $r=.1$, $(1.4,1.6) cap X = (1.4,1.6) subset [1,2]$.
Likewise, 1.0 is an interior point of $[1,2]$ because letting $r=.1$, $(0.9,1.1) cap X = (1.0,1.1) subset [1,2]$.
On the other hand, 3.5 is an exterior point because letting $r=.1$, $(3.4,3.6) cap X = emptyset$ .
All points in $X$ are either interior or exterior; X has no boundary points of $[1,2]$. So all of $[1,2]$‘s boundary points are outside $[1,2]$, a vacuous truth since there are no such points. So $[1,2]$ is relatively open in $X$.
Is that what’s happening, or is it something else?
You're right that there are no boundary points. In fact, every point is an interior point. Your argument that $3.5$ is an exterior point is incorrect because $(3.4,3.6)cap X=(3.4,3.6).$
Although the approach by @A.Sharma in the comments is the simpler, think it is still instructive for you to be able to prove it in your way.
Let's show every point is an interior point. Suppose $x in [1,2].$ Then $${y:|x-y|<1}cap X=[1,2) text{ or }[1,2] text{ or }(1,2].$$ In either case ${y:|x-y|<1}cap Xsubseteq X.$ Therefore every point of $[1,2]$ is an interior point and thus $[1,2]$ is open in $X.$
Similarly for $x in [3,4],$ $${y:|x-y|<1}cap X=[3,4) text{ or }[3,4] text{ or }(3,4].$$ In either case ${y:|x-y|<1}cap Xsubseteq X.$ Therefore every point of $[3,4]$ is an interior point and thus $[3,4]$ is open in $X.$
Answered by Sahiba Arora on November 2, 2021
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