It suffices to show that any proper subspace M of X is closed, since if M is not proper the result is trivial.  I am unsure how to approach this proof. Contradiction seems a little messy, as supposing M is not closed implies M is i.) open or ii.) neither closed nor open. Any help would be much appreciated.
121672
I know that probably this question has already been answered, but I'd like to present my attempt of solution.  Let $(E,\|\cdot\|)$ be a normed vector space and let $F\subseteq E$ be a finite-dimensional subspace. Suppose that $\dim_{\mathbb{R}} F=n$.  First of all I proved the following lemma.  Lemma 1 Every finite-dimensional normed vector space is complete.  Then, I noticed that $(F,\|\cdot\|_{F})$ is a finite-dimensional normed vector space, where $\|\cdot\|_{F}$ is the induced norm. So, $F$ is complete by Lemma 1.   Lemma 2 If  $(E,\|\cdot\|)$ is a normed vector space such that $\dim_{\mathbb{R}}{E}=n$, then $E$ is algebraically and topologically isomorphic to $\mathbb{R}^{n}$.  By Lemma 2, I can view $F$ as a complete subspace of $\mathbb{R}^n$ and, since $\mathbb{R}^n$ is itself complete, I can conclude that $F$ is closed. Here I used the fact that every complete subspace of a complete space is closed.  Is there something wrong?
78555
I was doing some linear algebra exercies. One I did is this: Prove that the union of two subspaces $U,W$ of $V$ is a subspace of $V$ if and only if one contains the other.   My proof is this: If one is a subset of the other $U \subset W$ then the union is $W$ which by assumption is a subspace. If neither is contained in the other there exists an example where the union is not a subspace: let $V$ be the xy-plane, $U$ the x-axis and $W$ the y-axis. Then the union is not closed with respect to addition.   My question is whether it is valid to prove something by example? How would you prove this in another way?
44045
Let $X$ be a banach space, and let $A$, and $B$ be closed linear subspaces. Assume that $$\inf\{\|x-y\|\mid x\in A, y\in B, \|x\|=\|y\|=1\}&gt;0$$  I want to show that $A+B$ is closed.   I was thinking of doing something like, let $z$ be a limit point of $A+B$, then there exist $z_n\in A+B$ such that $z_n\rightarrow z$, each $z_n$ can be written as $a_n+b_n$. Then I wanted to do something along the lines of determining whether $a_n$ and $b_n$ have a limit (if they do, then call them $a$ and $b$ and then $a\in A$, $b\in B$, and $z=a+b\in A+B$), but I can't seem to be able to use the condition given.   I am studying for a qual, so you can go ahead and either give a solution or sketch it.
32922
Let $X$ be topological space and $Y\subset X$. The goal is to show that $\dim Y\leq \dim X$.   Here we use this definition of dimension.  Let $Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_{n}$ be a chain of strictly increasing closed irreducible subsets of $X$ then $\dim X$ is the supremum over all such chains.   This is what I've got so far. I'm just stuck on one last bit. A hint would be nice.   Let $Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_n$ be a chain of irreducible closed subsets of $Y$. Since each $Y_i$ is irreducible, we just need to close them in $X$. There exists $F_i$ closed in $X$ such that $Y_i=F_i\cap Y$. Then if we closed each set in $X$, $\overline{F_0\cap Y}\subset\overline{F_1\cap Y}\subset \cdots \subset \overline{F_n\cap Y}$ is a chain of irreducible closed subsets of $X$. I just need to show that they are distinct. I really don't know where to proceed. A hint would be appreciated. Thanks.
59473
I wanted to find a quick proof of the following well-known fact. Since I couldn't easily find a reference, I provide a proof below.     Let $H$ be a separable Hilbert space, and $J\subset B(H)$ be a (not necessarily closed) two-sided ideal. If $T\in J$ is non-compact, then $J=B(H)$.    I'm also interested to know if there is a shorter/simpler argument.
1710
Let $X$ be a topological space and let $\left\{ Y_{i}\right\} _{i=1}^{\infty}$  be a descending chain of closed connected subsets of $X$. I know from reading elsewhere that ${\displaystyle \bigcap_{i=1}^{\infty}Y_{i}}$ is not necessarily a connected subspace of $X$   but I have no counter example and I haven't managed to come up with one.  There is a counter example  to the same question while also assuming $X$ is compact. However, it uses the quotient topology which I haven't studied about so I would prefer a different counterexample that does not use the quotient topology.  Help would be appreciated!
18595
Let $V$ be a Hausdorff topological vector space and $Z$ a linear subspace: $Z\leq X$  Is there a neat way to prove that: $$\dim Z\text{ finite}\implies Z\text{ closed}$$
50461
This could be classified as "homework", but I tried to solve this, made research online, and still failed, so I'll be glad to get some hints.  Let $G$ be a topological group, let $A$ be a compact subset of $G$, and let $B$ be a closed subset of $G$. Prove that $AB$ is closed.  If both $A$ and $B$ are not compact, but closed, this can fail, for example, if we let $A$ be the set of integers and $B$ the set of integer multiples of $\pi$, then both are closed, but $A+B$ is a proper dense subset of $\mathbb R$, so can't be closed. Also if $A$ is compact but $B$ is not closed, this easily fails.  Thanks
94728
How to show that quotient space $X/Y$ is complete when $X$ is Banach space, and $Y$ is a closed subspace of $X$?   Here's my attempt: Given a Cauchy sequence $\{q_n\}_{n \in \mathbb{N}}$ in $X/Y$, each $q_n$ is an equivalence class induced by $Y$, I want to find a representative $x_n$ in $q_n$ so that the induced sequence $\{x_n\}_{n \in \mathbb{N}}$ is also a Cauchy sequence in $X$. But I don't know how to construct such sequence.
89788
Let $M$ be the space of all $m\times n$ matrices. And $C=\{X\in M|\operatorname{rank}(X)\leq k\}$ where  $k\leq \min\{m,n\}$. Check whether the set $C$  is:   Closed Connected Compact Open   What are some other good properties of the set $C$,for example is it a manifold?  Clearly the set $C$ is closed  if someone is interested  a good proof can be found , hence $C$ is not open. Also as $C$ is unbounded therefore not compact. How to check whether the set $C$ is connected or not?
92686
Fact  The intersection of a closed set $A$ and a compact set $B$ is closed.  Background  I wanted to prove that the intersection of closed and compact sets is compact. There might be other ways, but I am not interested in them for now and am looking to prove the above statement.  Relevant Facts  I don't have much idea what to use. $A^c$ is open. Every cover $\cup_{i\ge1}U_i$ of $B$ contains a subcover $\cup_{i=1}^kU_i$. I might want to prove $A\cap B^c$ is open. Limit points may be another points to consider but I don't get any idea in that direction.  I also don't have that $B$ is closed otherwise it would be trivial.  Some Clarification  I am aware that compact sets are not closed in arbitrary topological spaces. The reason why I was motivated to believe tha intersection of closed and compact set is compact is true is: See Matt E.'s Answer      So if $K$ is compact in an arbitrary topological space (which is just to say that   it is a compact topological space when given its induced topology) and     $L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology   Is it true if $A$ and $B$ are not whole spaces?  I would appreciate any hints.
40144
Let $X$ be a subspace of a Hilbert space $H$.  prove:  $X$ is closed iff $X =( X^\perp)^\perp$   I do not know how to proceed. any hints would be appreciated.  thanks.  $X^\perp = ${ $y \in H$:$\langle x,y\rangle = 0 \forall x \in  X\}$.
69969
Let $(X,\tau)$ be a topological space, $U \subseteq X$ a non-empty open subset of $X$ and let $W$ be an irreducible non-empty closed subset of $X$. Assuming $W \cap U \neq \emptyset$ why is this intersection irreducible in $U$ and its closure (taken in $X$) equal $W$?  Clearly $W \cap U$ is dense in $W$ because it is an open subset of $W$ and $W$ is irreducible so its closure, taken in $W$ is $W$ itself. I don't see why its closure taken in $X$ is equal $W$. Why is this? Also, why is it irreducible in $U$?  EDIT: OK $W=cl_{X}(W)=cl_{X}(W \cap U)$ I think. Why is it irreducible in $U$ though?
79531
Let $X$ be the subspace $[0,1] \cup [2,3]$ of $\mathbb{R}$, and let $Y$ be the subspace $[0,2]$ of $\mathbb{R}$. The map $p \colon X \to Y$ defined by  $$ p(x) \colon=  \begin{cases} x \ &amp;\mbox{ for } \ x \in [0,1], \\ x-1 &amp;\mbox{ for } \ x \in [2,3]  \end{cases}  $$ is surjective and continuous. This much is clear to me.   But Munkres asserts that this map is also a closed (and hence a quotient) map. How to verify this?   Now if $A$ is the subspace $[0,1) \cup [2,3]$ of $X$, then the map $q \colon A \to Y$ obtained by restricting $p$ is continuous and surjective. Alright!   Now what stumps me is the following assertion by Munkres:   The set $[2,3]$ is saturated with respect to $q$. How to show this?
30304
I have a vector space $V$ over a field of characteristic $0$. If $M_1,\dots,M_k$ are proper subspaces of $V$, and $N$ is a subspace of $V$ such that $N\subseteq M_1\cup\cdots\cup M_k$, how can you tell $N\subseteq M_i$ for some $i$?  I was first trying to show it just in the case $N\subseteq M_1\cup M_2$, and hoping to extend it to finitely many $M_i$. If either of the $M_i$ contains the other, the claim follows, so I suppose neither $M_i$ contains the other.   In hopes of a contradiction, I suppose $N\not\subseteq M_1$ and $N\not\subseteq M_2$, Picking $x_1\in N\setminus M_1$ and $x_2\in N\setminus M_2$, I'd have $x_1+x_2\in N\subseteq M_1\cup M_2$. The only thing I could conclude was that actually $x_1,x_2\notin M_1$ and $x_1,x_2\notin M_2$, which seems like a dead end. What's the right way to do this?
27231
Let ($X,\tau)$ be a top space. Suppose $Y \subset X$ is a connected subspace given the subspace topology. If $A \cup B$ is a separation of the space $X$ then $Y$ lies entirely in $A$ or $B$. I just want a verification: Suppose other wise then $\exists a \in Y$ such that $a \in A$ and $\exists b \in Y$ such that $b \in B$. I claim that $$Y \cap X = Y \cap (A \cup B) = (Y \cap A) \cup (Y \cap B)$$ Forms a separation of $Y$. As $Y$ inherits the subspace topology, the topology on $Y$ can be given by $$\tau_Y = \{Y \cap U \mid U \in \tau\}$$ Then since $A,B \in \tau$ we have that $Y \cap A,Y \cap B \in \tau_Y$, i.e., are both open in $Y$. Clearly both are nonempty by existence of $a,b$ up top. Lastly for disjointness, suppose $\exists \alpha \in (Y \cap A) \cap (Y \cap B)$, then $\alpha \in A \cap B$ contradicting $A,B$ being disjoint (since they were assumed to form a separation of our parent space) and thus $$(Y \cap A) \cap (Y \cap B) = \emptyset$$ hence $$Y = (Y \cap A) \cup (Y \cap B)$$ forms a separation of $Y$ contradicting it being a connected subspace $\blacksquare$
66843
If $A$ is closed in $X$ and $B$ is closed in $Y$,   does it follow that $A \times B$ is closed in $X \times Y$ ?