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I'd like to start of by saying I am aware of this post, but I think my question is different enough to warrant its own post.

Some notation, because I don't think this is standard:

  • $N^X_\epsilon(x)$ is just the $\epsilon$-ball centered at $x$ in $X$. So say $X \subseteq \mathbb{R}$, we would have $N^X_\epsilon(x) = (x - \epsilon, x + \epsilon) \cap X$

  • "$\overset{\text{op}}{\subseteq}$" and "$\overset{\text{cl}}{\subseteq}$" means open and closed subsets respectively.

My lecturer supplied me with this proof of the unit interval being connected:

Let $U \subseteq I$ be an open and closed subset containing $0$. We want to show that $U = I$: if $q \in I$, we must show $q \in U$.

Since $U \overset{\text{cl}}{\subseteq} I \overset{\text{cl}}{\subseteq} \mathbb{R}$, we have that $U \overset{\text{cl}}{\subseteq} \mathbb{R}$. We get that $U \cap [0, q] \subseteq \mathbb{R}$ is non-empty, closed and bounded (by $1$), and thus has a maximum $z \in U \cap [0, q]$. If we show that $z = q$ we have succeded in showing $q \in U$.

Since $U \overset{\text{op}}{\subseteq} I$, there is an $\epsilon > 0$ such that $N_\epsilon^I(z) = (z - \epsilon, z + \epsilon) \cap I \subseteq U$, i.e., such that $$N^{[0, q]}_\epsilon(z) = (z - \epsilon, z + \epsilon) \cap [0, q] \subseteq U \cap [0, q]$$

I can follow the proof up until this part, but then he picks up his pace and suddenly says

The fact that $z$ is a maximum of $U \cap [0, q]$ then rules out the case when $z < q$; so $z = q. \quad \Box$

I fail to see why this last line is true, and what it has to do with the fact that $U$ is open in $I$. I think there's a lot happening between the lines here, and I wonder if anyone of you can explain exactly what's happening?

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$z$ is a max in $U \cap [0, q]$, and therefore is at least as great as the max of $U$ and at least as great as the max of $[0, q]$. Since the max of the latter is $q$, we have $z \ge q$.

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  • $\begingroup$ But we knew earlier in the proof that $U \cap [0, q]$ was non-empty, closed and bounded, so it had a maximum $z$. Couldn't we already draw that conclusion then? $\endgroup$ – kjQtte Nov 26 '16 at 18:01
  • $\begingroup$ I have no idea .. I didn't read the rest of the proof. What I did was explain the logic of the step you were confused about. :( $\endgroup$ – John Hughes Nov 26 '16 at 18:05
  • $\begingroup$ @JohnHughes Since $z$ is max in $U∩[0,q]$ we have $z\leq$ max $U$ and $z\leq q$. $\endgroup$ – Minato Jul 6 '18 at 15:27
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If $z<q$ than take any $z'$ such that $z<z'<q$ and $z'<z+\varepsilon$. Then $z'\in N^{[0,q]}_{ϵ}(z)=(z−ϵ,z+ϵ)∩[0,q]⊆U∩[0,q]$. So in $U∩[0,q]$ you find $z'$ that is greater than the max of $U∩[0,q]$.

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