It always possible to shrink a domain $U_\alpha$ to $V_\alpha$ in the sense that $$V_\alpha\Subset U_\alpha\quad\text{and} \quad\bigcup_{\alpha}U_\alpha=\bigcup_{\alpha}V_\alpha\,?$$   What does $\Subset$ mean ? Is it an error ?
120791
In some texts (mainly complex analysis or harmonic analysis) I sometimes see the following double subset symbol $\subset\subset$ for inclusion relation of two regions, e.g., $\Omega$ and $\Omega'$ are two regions in $\mathbb{C}$ such that $\Omega \subset\subset \Omega'$. I never figured out what it means exactly; I always interpreted it as the closure $\overline{\Omega}$ is contained in $\Omega'$ (so that some nasty boundary effects can be avoided). Is that right? Or does $\subset\subset$ mean some other kind of inclusion? Thanks.
75407
Exhibit an integral domain $R$ and a non-zero non-unit element of $R$ that is not a product of irreducibles.   My thoughts so far: I don't really have a clue. Could anyone direct me on how to think about this? I'm struggling to get my head round irreducibles.  Thanks.
7764
Why is the following true?  $$\{ \emptyset \}  \neq \emptyset $$  Is it just because the Empty set already refers to there being a set present and so it is just the same argument that $\{ \{3\} \} \neq \{ 3\} $?  And is there a proof for the following?   For any set$\, X$,   $$X \cup \emptyset = X \quad \text{and} \quad X \cap \emptyset = \emptyset$$
18725
Let $X$ be a compact metric space and $Y$ any metric space.  If $f:X \to Y$ is continuous, then $f(X)$ is compact (that is, continuous functions carry compact sets into compact sets). Proof:  Consider an open cover of $f(X)$. Then $f(X) \subset \bigcup_{\alpha \in A}V_\alpha$ where each $V_\alpha$ is open in $Y$. $X \subset f^{-1}(f(X)) \subset f^{-1}\left(\bigcup_{\alpha \in A}V_\alpha\right) = \bigcup_{\alpha \in A}f^{-1}(V_\alpha)$. Hence $\bigcup_{\alpha \in A}f^{-1}(V_\alpha)$ is an open cover of $X$.  Since $X$ is compact then we can choose a finite subcover $\{V_i\}_{i=1}^n$ such that $X \subset \bigcup_{i=1}^n f^{-1}(V_i)$. So then $f(X) \subset f\left(\bigcup_{i=1}^n f^{-1}(V_i)\right) = \bigcup_{i=1}^n f\left(f^{-1}(V_i)\right) \subset \bigcup_{i=1}^n V_i$, a finite subcover of $f(X)$.  $\therefore f(X)$ is compact.  Does this proof have an error? Thanks for your help.
48529
Could anyone provide an example of a nonempty subset $U$ of $R^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $R^2$
65673
Basically does a vector set that is linearly independent in $R^n$ automatically span $R^n$?  My initial thought is yes, but is there some counterexample that can disprove this?
117117
I have the following topology : $$\tau= \Bigl\{U\subseteq \mathbb{R}^2: (\forall(a,b) \in U) (\exists  \epsilon &gt;0) \bigl([a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq U\bigr)\Bigr\}$$  Are these a basis for the previous topology:  $\beta_1= \{[a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon&gt;0 \}$  $\beta_2= \{[a,a+\epsilon) \times [b-\epsilon, b+\epsilon)\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon&gt;0 \}$  The first one is obviously a basis for $\tau$ because of the definition of $\tau$ and I would say that the second is also a basis, because $[a,a+\epsilon) \times [b-\epsilon, b+\epsilon) \subseteq [a,a+\epsilon] \times [b-\epsilon, b+\epsilon]  $  Is it correct? what do you think?
117875
I have been trying to prove the following for a while now, but I can't get to the solution:  $$(A\times C)\cup(B\times D)\subseteq(A\cup B)\times(C\cup D)$$  Can anyone help me?
9688
Is this true?  If $A$ and $B$ are sets, then $A=\emptyset $ if and only if $B = A \bigtriangleup B$.  If $A=\emptyset$ then $B=\emptyset$ too?  Could someone help me please?
45166
I am here to ask something that I think it is interesting, first I just read about the shrinkage using the Ridge or Lasso regression by using the lambda as the penalty to introduce a little bias that reduces a large amount of variance. From here why lambda is doing is it flatter the slope that makes me think what if it does the opposite thing. It increases the slope by change the lambda &lt; 0, I know we don't do that in shrinkage so my question is can it be beneficial for lambda &lt; 0 in some expansion instead of shrinkage?. Are there any case it can be applied?
65710
In a Linear Algebra textbook I am reading, the following is stated: $b\notin \operatorname{span}(A \cup \{a\})\setminus \operatorname{span}(A)$. It does so without explaining what "$\setminus$" means. I apologize if this question does not belong here but I just want to understand what it means. I can close the question if someone just comments on its meaning.
26905
Assume $\bar A\cap\bar B=\emptyset$. Is $\partial (A \cup B)=\partial A\cup\partial B$, where $\partial A$ and $\bar A$ mean the boundary set and closure of set $A$?   I can prove that $\partial (A \cup B)\subset \partial A\cup\partial B$ but for proving $\partial A\cup\partial B\subset \partial (A \cup B)$ it seems not trivial. I tried to show that for $x\in \partial A\cup\partial B$ WLOG, $x\in \partial A$ so $B(x)\cap A$ and $B(x)\cap A^c$ not equal to $\emptyset$ but it seems not enough to show the result.
114169
Why does $\bigcup \emptyset = \emptyset$ but $\bigcap \emptyset$ is not defined?    If I had to guess, I'd say it's also equal to $\emptyset$.
70126
Does a set of $n$ independent vectors in $R^{n}$ always span  $R^{n}$?  If so, what's the reasoning behind this?
101843
Take for example the set $X=\{a, b\}$. I don't see $\emptyset$ anywhere in $X$, so how can it be a subset?
57226
I'm working through a book on fundamental topology, and I'm stuck at a proof in set theory. The book provides a proof for:   $$\text{Let }\left\{A_\alpha\right\}_{\alpha\in I}\text{ be an indexed family of subsets of a set }S\text{; then:}$$ $$C\left(\bigcup_{\alpha\in I}A_\alpha\right)=\bigcap_{\alpha\in I}C\left(A_\alpha\right)$$  where $I$ is an indexing set for $\alpha$, and $\alpha\in I$, and $C(S)$ denotes the complement set. They do this by proving first that, if $x\in C\left(\bigcup_{\alpha\in I}A_\alpha\right)$, then $C\left(\bigcup_{\alpha\in I}A_\alpha\right)\subset\bigcap_{\alpha\in I}C\left(A_\alpha\right)$. Then, the prove the converse; that if $x\in\bigcap_{\alpha\in I}C\left(A_\alpha\right)$, then $\bigcap_{\alpha\in I}C\left(A_\alpha\right)\subset C\left(\bigcup_{\alpha\in I}A_\alpha\right)$, so the two sides, being subsets of each other, must be equal.  That makes perfect sense, and is very clear and straightforward. However, what I don't understand is their second proof (under the same conditions):  $$C\left(\bigcap_{\alpha\in I}A_\alpha\right)=\bigcup_{\alpha\in I}C\left(A_\alpha\right)$$  While this statement intuitively makes sense to me, I can't seem to prove it mathematically. Here's what I've done so far:  Let $x\in C\left(\bigcap_{\alpha\in I}A_\alpha\right)$. We know that, conversely, $x\notin \bigcap_{\alpha\in I}A_\alpha$. This is where the proofs diverge, though; if it were the first formula, it would be possible to say that $x\notin \bigcup_{\alpha\in I}A_\alpha$, and as a result, that $x\notin \left\{A_\beta\right\}_{\beta\in I}$. However, I can no longer do that, and approaching it from the other direction gives the same result.  Let $x\in\bigcup_{\alpha\in I}C\left(A_\alpha\right)$. We know that $x\notin\bigcup_{\alpha\in\ I}A_\alpha$, but this still doesn't get me anywhere.   How should I approach this proof? Any hints or pointers for someone new to topology?
42854
Please let me know if you find my proof satisfactory.  Let $A\subset B\subset \mathbb{R}^n$, where $\bar{A}$ is the closure of $A$.  If $\vec{x}\in \bar{A}$ then $\vec{x}\in A^\circ$ or $\vec{x}\in ∂A$ (in the boundary of $A$). If $\vec{x}\in A^\circ$ then $\vec{x}\in A \implies \vec{x}\in B\subset \bar{B}$. If $\vec{x} \in ∂A$ then $B_\varepsilon(\vec{x})\cap A\ne \emptyset \ne B_\varepsilon(\vec{x})\cap B$ (for any $\varepsilon &gt; 0$). Thus, since $A\subset B$, $\vec{x}\in B \implies \bar{A}\subset \bar{B}$.