Show that the projection map of $[0,1]  \times \mathbb{R} $ onto $\mathbb{R}$ is a closed map.   I know that [0,1] is a compact space but I am not sure how this implies that the projection map is closed. I know that in plain old $\mathbb{R} \times \mathbb{R} $ the projection map is not closed because the graph of $f(x) = 1/x $ is closed but has open projection and that in $[0,1]  \times \mathbb{R} $ we can't have this type of function, but that's it. Thanks for any tips or help.
64644
Let $\pi: X \times Y \to X$ be the projection map where $Y$ is compact. Prove that $\pi$ is a closed map.   First I would like to see a proof of this claim.  I want to know that here why compactness is necessary or do we have any other weaker condition other than compactness for the same result to hold.
59414
How would you show that there is no continuous injective map  $f:\mathbb{R}^3 \to \mathbb{R}$?  I tried the approach where if $a \in \mathbb{R}$ then $\mathbb{R} \backslash \{a\} $ is not clearly connected so there can't exist a continuous function otherwise $\mathbb{R} \backslash \{a\} $ would be connected?
98576
My proof is very different from my reference, hence I am wondering is I got this right?  Apparently, $F$ is continuous, and the identity matrix is closed. Now we want to show that the preimage of continuous function on closed set is closed.  Let $D$ be a closed set, Consider a sequence $x_n \to x_0$ in which $x_n \in f^{-1}(D)$, and we will show that $x_0 \in f^{-1}(D)$.  Since $f$ is continuous, we have a convergent sequence $$\lim_{n\to \infty} f(x_n) = f(x_0) = y.$$  But we know $y$ is in the range, hence, $x_0$ is in the domain. So the preimage is also closed since it contains all the limit points.  Thank you.
51499
Let $T: X \longrightarrow Y$ be a continuous linear map between two Banach spaces.  When is $\operatorname{Ran}(T)$ a closed subspace?  What theorems are there?   Thanks :)
17119
I want to show that $\mathbb{R}^n\times\{0\}$ has measure zero in $\mathbb{R}^{n+1}$.  For example, take $n=1$. I want to show that the $x$-axis has measure zero in the plane. I cover it with the sets $[-1,1]\times[-\epsilon/8,\epsilon/8]$, $[-2,2]\times[-\epsilon/32,\epsilon/32]$, $\ldots$  The goal is to have the volumes be $\epsilon/2, \epsilon/4, \ldots$ so that they sum to $\epsilon$.  I think this method generalizes to $\mathbb{R}^n$, simply by choosing $[-1,1]^{n-1}\times[-\epsilon/2^{n+2},\epsilon/2^{n+2}],\ldots$. I don't think it is very elegant though. Is there a "better" way to do this?
102971
Let $GL(n,\mathbb{R})$ be the space of invertible $n \times n$ matrices over $\mathbb{R}$. I have already proved that $GL(n,\mathbb{R})$ is open in $\mathbb{R}^{n^2}$ and that it is not connected with the relative topology from $\mathbb{R}^{n^2}$. How do I prove that the map $f:GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$ given by $f(x):=x^{-1}$ is continuous?  Any help would be greatly apprecaited.   Additionally how do I prove that $g:GL(n,\mathbb{R})\times GL(n,\mathbb{R}) \rightarrow GL(n,\mathbb{R})$ given by $g(x,y):=xy$ (matrix multiplication) is continuous, where we take the relative topology from $\mathbb{R}^{n^2} \times \mathbb{R}^{n^2} \cong \mathbb{R}^{2n^2}$?
11379
A continuous map from a compact space to a Hausdorff space is closed. Why this is true?  Help me please I want to learn why this is correct.
89772
Can someone give me an example of noncomplete inner product space $H$, its closed linear subspace of $H_0$ and element $x\in H$ such that there is no orthogonal projection of $x$ on $H_0$. In other words I need to construct a counterexample to theorem about orthogonal projection when inner product space is not complete.
41934
Let $f:A\to C$ and $g:B\to D$ be bounded linear maps of Banach spaces with closed image. Will $f\widehat{\otimes}g:A\widehat{\otimes}_\pi B\to C\widehat{\otimes}_\pi D$ also have closed image? What about injective tensor products?
67090
Question is :  Suppose  $f: [0,1]\rightarrow (0,1)$ is Continuous then which of the following is NOT true..   $F\subseteq[0,1]$ is closed set implies $f(F)$ is closed in $\mathbb{R}$ If $f(0)&lt;f(1)$ then $f([0,1])$ must be equal to $[f(0),f(1)]$ There must exist $x\in(0,1)$ such that $f(x)=x$ $f([0,1])\neq (0,1)$   Continuous map need not map closed sets to closed sets..  So, first option is not true...  Continuous maps takes connected sets to connected sets ...  So $f([0,1])$ must be connected and it is equal to  $[f(0),f(1)]$.. So, Second option is true..  Continuous maps takes compact sets to compact sets...  So, $f([0,1])\neq (0,1)$ and so fourth option is true...  I guess third option is also false though I can not think of any example..  Continuous map from compact set to itself has a fixed point.  But how do i conclude that this would imply third option is true/false.  Please help me to clear this...  Thank you.
60920
How to prove that a compact set $K$ in a Hausdorff topological space $\mathbb{X}$ is closed? I seek a proof that is as self contained as possible.  Thank you.
14317
I'm trying to solve this question:  Let $$M=\{f\in C([0,1])\ | \ f(1)=0\}.  $$  Is $M$ closed in $(C[0,1]),d_1)$?  At first, I assumed that $M$ was closed, so I tried to prove that if there is a sequence $f_1,f_2,f_3,...\in M,$and $\ f_n\rightarrow f\,$ with respect to $d_1$, then $f \in M$.    However, I got stuck and checked out the answers, and the answer I was given was just a counterexample of a sequence $(f_n)_n$ where $f_n \in M,\ f_n\rightarrow f$ wrt $d_1$, and $f\notin M$.    I understand that it's a lot easier to give a counterexample, but unfortunately I suck at analysis and cannot tell whether a set is closed or not upon inspection or come up with a counterexample right away.  Are there other ways of proof, like proof by contradiction, to show that this set is not closed?
53345
Proposition : The only sets in $R^n$ which are both open and closed are the empty set and $R^n$ itself.  I came up with the proof of this claim and I'd like to know if my proof is correct.  Proof:  Let $S$ be the nonempty proper subset of $R^n$ which is both open and closed. Then $T = R^n-S$ is also a nonempty proper subset of $R^n$ which is both open and closed. Now take a point $s_0$ in $S$ and $t_0$ in $T$. Then we can consider a closed box $J_1$ which contains both $s_0$ and $t_0$.  The n-dimensional interval $I_1^k$ is defined by the inequalities   $s_k\le x_k \le t_k$ (k=1, 2, $...$, n ), where each $s_k$ and $t_k$ are the k-th coordinates of $s_0$ and $t_0$. Without loss of generality, assume that for all k, $s_k \le t_k$.  Here $J_1$ denotes the catesian product   $J_1$ = $I_1^1$ x  $\cdots$ x $I_n^1$   Now, bisect all of the n 1-dimensional intervals and the point $\frac {s_0+t_0}{2}$ will lie in the center of the bisected box. This point belongs to either $S$ or $T$. Say it belongs to $S$ then, rename it $s_1$ and the original $t_0$ to be $t_1$. Now again consider a box that has both points as vertices like the previous case and repeat the process.   Writing $I_k^m$ = [$s_k^m$,$t_k^m$],   we have  $t_k^m$ - $s_k^m$ = $\frac {t_k-s_k}{2^{m-2}}$  Then for each fixed k, the sup of all left endpoints must therefore be equal to the inf of all right endpoints, and their common value we donote by $b_k$ (actually we know this by Bolzano-Weierstrass).   Now we assert that b=($b_1$,$...$,$b_n$) is a limit point of $S$ and $T$. To see this, take any n-ball B(b; $r$). The point b of course belongs to each of the boxes $J_1$, $J_2$, $...$ constructed above, and when $m$ is such that $max_k\frac {t_k-s_k}{2^{m-2}}$ &lt; r/2, this neighborhood will include $J_m$ and since $J_m$ contains both points of $S$ and $T$, b is a limit point of both sets. But since both sets are closed, they both contain b, which is a contradiction. Hence, the proposition is true.
104103
I'm trying to prove that proper subspaces of $ R^n $ are closed and have empty interiors.  To prove that they are closed I'm trying to use the fact that inverse images of closed sets are closed sets by continuous functions. So I'm trying to find a linear map such that a proper subspace of $ R^n $ is its Kernel but I don't know how to express it so that it would work on all dimensions.  Regarding the fact that they have empty interiors, I really don't know what to do.  Thanks in advance.
5150
Let $X=\mathbb{R}$ and $S=[-1,1]$, i don't quite understand why $[-1,0)$ is open in $S$ but not open in $X$. As far as i know, to show a set is open, we need to show there exist an open ball $B(x,r), r&gt; 0$ st for all $x\in [-1,0)$ ,$x\in B\subset S$, however, i still can't see why  $[-1,0)$ is open in $S$ but not open in $X$. When we look at -1, i don't think this is a interior point of S.
77246
Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be a continuously differentiable map satisfying $\|f(x)-f(y)\|\geq\|x-y\|$ for all $x,y\in\mathbb{R}^{n}.$ Then   $A.$ $f$ is onto.  $B.$ $f(\mathbb{R}^{n})$ is a closed subset of $\mathbb{R}^{n}.$  $C.$ $f(\mathbb{R}^{n})$ is a open subset of $\mathbb{R}^{n}.$  $D.$ $f(0)=0.$  According to me $f$ is one one and onto so image is closed and open both. Please suggest me. Thanks in advance.
80122
I have troubles showing the following:  Let $(X,\rho)$ be a compact metric space and $F \subset X$ a closed subset. Prove that if diam $F &lt; \infty$, then there exist $x_{0}, y_{0} \in F$ such that diam $F= \rho(x_{0},y_{0})$.  It looks so trivial yet hard to find the trick.