5
votes
Understanding the witness property in the Henkin construction
You're correct that the two versions are equivalent when $T$ is complete (maximal), and your proposed argument (alternating between completing the theory and adding witnesses) works just fine.
In fact,...
- 80.1k
3
votes
Accepted
Heine-Borel for Finite Dimensional Normed Vector Spaces
Throughout, let $\Bbb{F}$ be either $\Bbb{R}$ or $\Bbb{C}$. Recall that if $X,Y$ are normed vector spaces over $\Bbb{F}$ and $T:X\to Y$ is linear, then the following statements are equivalent:
$T$ is ...
- 60.2k
2
votes
Accepted
Pointwise convergence of almost periodic function
Put $K=\overline{\{f_t\}_{t\in \mathbb{R}}}^{\|\cdot\|_{\infty}}$. Since the set $K$ is compact in the given set $X$ of functions from $\mathbb R$ to $\mathbb R$, with the topology $\tau_\infty$, ...
- 95.4k
2
votes
Is $X\simeq [0,1]$?
I'm not sure why, but this question was answered and then the answer was deleted less than an hour later. I will post this answer again, but if the original answerer undeletes and lets me know, then I'...
- 104k
1
vote
Locally Lipschitz with respect to a variable uniformly to another implies Lipschitz for every compact subset
Here is a counterexample if $A$ is disconnected:
Let $A_p=\mathbb{R} \times B(p,1)$, and $A = A_{-2} \cup A_2$. Define
$f(t,x) = \begin{cases} 0, & (t,x) \in A, t \le 0 \\
0, & (t,x) \in A_{-2}...
- 175k
1
vote
Why do we need step 4. in this proof of the compactness of $[0, 1]$?
In your question you restate Spivak's proof. It seems that you already found the answer to Q1. However, your formulation of $(4)$ is misleading:
There is an open interval $J$, $\alpha\in J\subseteq U$...
- 4,749
1
vote
Accepted
Why do we need step 4. in this proof of the compactness of $[0, 1]$?
The purpose of step 4 is to provide an interval which contains $\alpha$ and some point $x \in A$. As you say, $U$ may be made of disconnected intervals, so we need to first pick an interval within $U$ ...
- 1,253
1
vote
Accepted
Quasicompactness of $\operatorname{Proj}A$
I have figured this out.
By hypothesis we have that:
$$\sqrt{A_+}=\sqrt{\langle f_1,\dots, f_n\rangle}$$
hence:
$$V(A_+)=\bigcap_{i=1}^nV(f_i)$$
By taking compliments, it follows that:
$$\operatorname{...
- 3,994
1
vote
Accepted
If $(X,\tau)$ is compact and $\mathcal{F}\subset C(X)$ is separating points, then the weak topology generated by $\mathcal{F}$ is equal to $\tau$
As $\sigma\subset \tau$, you know that the identity map $1_X\colon(X,\tau)\to (X,\sigma)$ is continuous. Clearly it is a bijection.
You are given that $(X,\tau)$ is compact.
$(X,\sigma)$ is Hausdorff,...
- 13.2k
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