Abstract Nonsense

Crushing one theorem at a time

Munkres Chapter 2 Section 18


Problem: Show that the normal \varepsilon-\delta formulation of continuity is equivalent to the open set version.

Proof: Suppose that \left(\mathcal{M},d\right),\left(\mathcal{N},d'\right) are metric spaces and for every \varepsilon>0 and every f(x)\in\mathcal{N} there exists some \delta>0 such that f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x)). Then, given an open set U\subseteq \mathcal{N} we have that f^{-1}(U). To see this let x\in f^{-1}(U) then f(x)\in U and since U is open by hypothesis there exists some open ball B_\varepsilon(f(x)) such that B_{\varepsilon}(f(x))\subseteq U and thus by assumption of \varepsilon-\delta continuity there is some \delta>0 such that \displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq U and so B_{\delta}\subseteq f^{-1}(U) and thus x is an interior point of f^{-1}(U).

Conversely, suppose that the preimage of an open set is always open and let f(x)\in\mathcal{N} and \varepsilon>0 be given. Clearly B_{\varepsilon}(f(x)) is open and thus f^{-1}\left(B_{\varepsilon}(f(x))\right) is open. So, since x\in f^{-1}\left(B_{\varepsilon}(f(x))\right) there exists some \delta>0 such that B_{\delta}(x)\subseteq f^{-1}\left(B_{\varepsilon}(f(x))\right) and so

f\left(B_{\delta}(x)\right)\subseteq f\left(f^{-1}\left(B_{\varepsilon}(f(x))\right)\right)\subseteq B_{\varepsilon}(f(x))



Problem: Suppose that f:X\to Y is continuous. If x is a limit point of the subset A of X, is it necessarily true that f(x) is a limit point of f(A)?

Proof: No. Consider (-1,0)\cup(0,1) with the suspace topology inherited from \mathbb{R} with the usual topology. Define

f:(-1,0)\cup(0,1)\to D:x\mapsto\begin{cases}0\quad\text{if}\quad x\in(-1,0)\\ 1\quad\text{if}\quad x\in(0,1)\end{cases}

This is clearly continuous since f^{-1}(\{1\})=(-1,0) and f^{-1}(\{1\})=(0,1) which are obviously open. But, notice that \frac{-1}{2} is a limit point for (-1,0) since given a neighborhood N of \frac{-1}{2} we must have that there is some (a,b)\cap \left((-1,0)\cup(0,1)\right)\cap \subseteq N which contains it. But, f\left(\frac{-1}{2}\right)=\{0\} is not a limit point for f\left((-1,0)\right)=\{0\} since that set has no limit points. \blacksquare


Problem: Let X and X' denote a singlet set in the two topologies \mathfrak{J} and \mathfrak{J}' respectively. Let \text{id}:X'\to X be the identity function. Show that

a) \text{id} is continuous if and only if \mathfrak{J}' is finer than \mathfrak{J}

b) \text{id} is a homeomorphism if and only if \mathfrak{J}=\mathfrak{J}'


a) Assume that \text{id} is continuous then given U\in\mathfrak{J} we have that \text{id}^{-1}(U)=U\in\mathfrak{J}'. Conversely, if \mathfrak{J}' is finer than \mathfrak{J} we have that given U\in\mathfrak{J} that \text{id}^{-1}(U)=U\in\mathfrak{J}'

b) If \text{id} is a homeomorphism we see that both it and \text{id}^{-1}=\text{id}:X\to X' are continuous and so mimicking the last argument we see that \mathfrak{J}\subseteq\mathfrak{J}' and \mathfrak{J}'\subseteq\mathfrak{J}. Conversely, if \mathfrak{J}=\mathfrak{J}' then we now that

U\in\mathfrak{J}\text{ iff }U\in\mathfrak{J}' or equivalently that U\text{ is open in }X\text{ iff  }\text{id}(U)=U\text{ is open in }X'

which defines the homeomorphic property. \blacksquare


Problem: Given x_0\in X and y_0\in Y show that the maps f:X\to X\times Y and g:X\times Y\to Y given by f:x\mapsto (x,y_0) and g:y\mapsto (x_0,y) are topological embeddings.

Proof: Clearly f and g are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, f(x)=(x,y_0)=(x',y_0)=f(x') then by definition of an ordered pair we must have that x=x'.  Lastly, the inverse function is continuous since f^{-1}:X\times \{y_0\}\to X:(x,y_0)\mapsto x is the restriction of the projection to X\times\{y_0\}. The same is true for g. \blacksquare


Problem: Show that with the usual subspace topology [0,1]\approx[a,b] and (0,1)\approx(a,b).

Proof: Define f:[0,1]\to[a,b]:x\mapsto (b-a)+a and g:(0,1)\to(a,b):x\mapsto (b-a)+a. These are easily both proven to be homeomorphisms. \blacksquare


Problem: Find a function f:\mathbb{R}\to\mathbb{R} which is continuous at precisely one point.

Proof: Define

f:\mathbb{R}\to\mathbb{R}:\begin{cases}x\quad\text{if}\quad x\in\mathbb{Q}\\ 0\quad\text{if}\quad x\notin\mathbb{Q}\end{cases}

Suppose that f is continuous at x_0, then choosing sequences \{q_n\}_{n\in\mathbb{N}},\{i_n\}_{n\in\mathbb{N}} of rational and irrationals numbers respectively both converging to x_0. We see by the limit formulation of metric space continuity that

x_0=\lim\text{ }q_n=\lim\text{ }f(q_n)=f(x_0)=\lim\text{ }f(i_n)=\lim\text{ }0=0

And so if f were to be continuous anywhere it would have to be at 0. To show that it is in fact continuous at 0 we let \varepsilon>0 be given then choosing \delta=\varepsilon we see that |x|<\delta\implies |f(x)|\leqslant |x|<\delta=\varepsilon from where the conclusion follows since this implies that \displaystyle \lim_{x\to 0}f(x)=0=f(0). \blacksquare



a) Suppose that f:\mathbb{R}\to\mathbb{R} is “continuous from the right”, that is, \displaystyle \lim_{x\to a^+}f(x)=f(a) for each a\in\mathbb{R}. Show that f is continuous when considered as a function from \mathbb{R}_\ell to \mathbb{R}.

b) Can you conjecture what kind of functions f:\mathbb{R}\to\mathbb{R} are continuous when considered as maps as \mathbb{R}\to\mathbb{R}_\ell. As maps from \mathbb{R}_\ell to \mathbb{R}_\ell?


a) Note that by the assumption that \displaystyle \lim_{x\to a^+}f(x)=f(a) we know that for every \varepsilon>0 there exists some \delta>0 such that 0\leqslant x-a<\delta implies that |f(x)-f(a)|<\varepsilon. So, let U\subseteq\mathbb{R} be open and let a\in f^{-1}(U). Then, f(a)\in U and since U is open we see that there is some \varepsilon>0 such that B_{\varepsilon}(f(a))\subseteq U. But, by assumption there exists some \delta>0 such that 0\leqslant x-a<\delta\implies f(x)\in B_{\varepsilon}(f(a)). But, \left\{x: 0\leqslant x-a<\delta\right\}=[a,a+\delta) and thus f\left([a,a+\delta)\right)\subseteq B_{\varepsilon}(f(a))\subseteq U and thus [a,a+\delta)\subseteq f^{-1}(U) and so a is an interior point for f^{-1}(U) from where it follows that f^{-1}(U) is open and thus f is continuous.

b) I’m not too sure, and not too concerned right now. My initial impression is that if f:\mathbb{R}\to\mathbb{R}_\ell is continuous then f^{-1}([a,b)) is open which should be hard to do. Etc.


Problem: Let Y be an ordered set in the order topology. Let f,g:X\to Y be continuous.

a) Show that the set \Omega=\left\{x\in X:f(x)\leqslant g(x)\right\} is closed in X

b) Let h:X\to Y:x\mapsto \max\{f(x),g(x)\}. Show that h is continuous.


a) Let x_0\notin\Omega then f(x_0)>g(x_0). Suppose first that there is no g(x_0)<\xi<f(x_0) and consider

f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left((-\infty,f(x_0)\right)=U

This is clearly open in X by the continuity of f,g and x_0 is contained in it. Now, to show that U\cap \Omega=\varnothing let z\in U then f(z)\in f\left(f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left(-\infty,f(x_0)\right)\right) which with simplification gives the important part that f(z)\in (g(x_0),\infty) and so f(z)>g(x_0) but since there is no \xi such that g(x_0)<\xi<f(x_0) this implies that f(z)\geqslant f(x_0). Similar analysis shows that g(z)\in (-\infty,f(x_0)) and since there is no \xi as was mentioned above this implies that g(z)\leqslant g(x_0). Thus, g(z)\leqslant g(x_0)<f(x_0)\leqslant f(z) and thus z\notin\Omega.

Now, suppose that there is some \xi such that g(x_0)<\xi<f(x_0) then letting V=f^{-1}(\xi,\infty)\cap g^{-1}(-\infty,\xi) we once again see that V is open and x_0\in V. Furthermore, a quick check shows that if z\in V that f(z)\in(\xi,\infty) and so f(z)>\xi and g(z)\in(-\infty,\xi) and so g(z)<\xi and so f(z)>g(z) so that z\notin\Omega. The conclusion follows

b) Let \Omega_f=\left\{x\in X:f(x)\geqslant g(x)\right\} and \Omega_g=\left\{x\in X:g(x)\geqslant f(x)\right\}. As was shown in a) both \Omega_f,\Omega_g are closed and thus define

f\sqcup g:X=\left(\Omega_f\cup\Omega_g\right)\to Y:x\mapsto\begin{cases}f(x)\quad\text{if}\quad x\in\Omega_f\\ g(x)\quad\text{if}\quad x\in\Omega_g\end{cases}

Notice that since f,g are both assumed continuous and f\mid_{\Omega_g\cap\Omega_f}=g\mid_{\Omega_f\cap\Omega_g} that we may conclude by the gluing lemma that f\sqcup g is in fact continuous. But, it is fairly easy to see that f\sqcup g=\max\{f(x),g(x)\} \blacksquare


Problem: Let \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of subset of X; let \displaystyle X=\bigcup_{\alpha\in\mathcal{A}}U_\alpha. Let f:X\to Y and suppose that f\mid_{U_\alpha} is continuous for each \alpha\in\mathcal{A}

a) Show that if the collection \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is finite each set U_\alpha is closed, then f is continuous.

b) Find an example where the collection \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is countable and each U_\alpha is closed but f is not continuous.

c) An indexed family of sets \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is said to be locally finite if each point of X has a neighborhood that intersects only finitely many elements of \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}. Show that if the family \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is locally finite and each U_\alpha is closed then f is continuous.$


a) This follows since if V\subseteq Y is closed then it is relatively easy to check that \displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V) but since each f\mid_{U_\alpha} is continuous we see that \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in U_\alpha. But, since each U_\alpha is closed in X it follows that each \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in X. Thus, f^{-1}(V) is the finite union of closed sets in X, and thus closed.

b) Give [0,1] the subspace topology inherited from \mathbb{R} with the usual topology and consider \left\{f_n\right\}_{n\in\mathbb{N}-\{1,2\}} with

f_n=\iota_{[0,1-\frac{1}{n}]}:\left[0,1-\tfrac{1}{n}\right]\to[0,1]:x\mapsto x

Clearly each f_n i

Lemma: Let Y be any topological space and \left\{V_\beta\right\}_{\beta\in\mathcal{B}} be a locally finite collection of subsets of Y. Then, \displaystyle \bigcup_{\beta\in\mathcal{B}}\overline{V_\beta}=\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}

Proof: The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let \displaystyle x\in\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta} since the collection of sets is locally finite there exists some neighborhood N of x such that it intersects only finitely many, say V_{\beta_1},\cdots,V_{\beta_n}, elements of the collection. So, suppose that x\notin \left(\overline{V_{\beta_1}}\cup\cdots\cup \overline{V_{\beta_n}}\right) then N\cap-\left(\overline{V_{\beta_1}}\cup\cdots\cup\overline{V_{\beta_n}}\right) is a neighborhood of x which does not intersect \displaystyle \bigcup_{\beta\in\mathcal{B}}V_\beta contradicting the assumption it is in the closure of that set. \blacksquare

Now, once again we let V\subseteq Y be closed and note that \displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V) and each \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in U_\alpha and since U_\alpha is closed in X we see that \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in X. So, noting that \left(f\mid_{U_\alpha}\right)^{-1}(V)\subseteq U_\alpha it is evident from the assumption that \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is locally finite in X that so is \left\{\left(f\mid_{U_\alpha}\right)^{-1}(V)\right\}_{\alpha\in\mathcal{A}} and thus (for notational convenience) letting F_\alpha=\left(f\mid_{U_\alpha}\right)^{-1}(V) the above lemma implies that

\displaystyle \overline{f^{-1}(V)}=\overline{\bigcup_{\alpha\in\mathcal{A}}F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}\overline{F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}F_\alpha=f^{-1}(V)

From where it follows that the preimage of a closed set under f is closed. The conclusion follows. \blacksquare


Problem: Let f:A\to B and g:C\to D be continuous functions. Let us define a map f\times g:A\times C\to B\times D by the equation (f\times g)(a\times c)=f(a)\times g(c). Show that f\times g is continuous.

Proof: This follows from noting the two projections of f\times g are \pi_1\circ(f\times g):A\times B\to C:a\times b\mapsto f(a) and \pi_2\circ(f\times g):A\times B\to D:a\times b\mapsto f(b). But, both of these are continuous since \left(\pi_1(f\times g)\right)^{-1}(U)=f^{-1}(U)\times B. To see this we note that x\in f^{-1}(U)\times B if and only if x\in f^{-1}(U) which is true if and only if f(x)=\left(\pi_1\circ(f\times g)\right)(x)\in U or in other words x\in \left(\pi_1\circ(f\times g)\right)^{-1}(U). Using this we note that the preimage an open set in C will be the product of open sets by the continuity of f. It clearly follows both projections, and thus the function itself are continuous. \blacksquare


Problem: Let F:X\times Y\to Z. We say that F is continuous in eahc variable separately if for each y_0\in Y, the map h:X\to Z:x\mapsto F(x\times y_0( is continuous and for each x_0\in X the map j:Y\to Z:y\mapsto F(x_0\times y) is continuous. Show that if F is continuous then F is continuous in each variable separately.

Proof: If F is continuous then clearly it is continuous in each variable since if we denote by G_{y_0} the mapping G_{y_0}:X\to Z:x\mapsto F(x\times y_0) we see that G_{y_0}=H_{y_0}\circ(F\mid_{X\times\{y_0\}}) where H_{y_0}:X\to X\times Y:x\mapsto x\times y_0 but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable.


Problem: Let F:\mathbb{R}\times\mathbb{R}\to\mathbb{R} be given by

\displaystyle F(x\times y)=\begin{cases} \frac{xy}{x^2+y^2}&\mbox{if}\quad x\times y\ne 0\times 0\\ 0 &\mbox{if} \quad x\times y=0\times0\end{cases}

a) Show that F is continuous in each variable separately.

b) Compute g:\mathbb{R}\to\mathbb{R}:x\mapsto F(x\times x).

c) Show that F is not continuous


a) Clearly both F(x\times y_0) and F(x_0\times y) are continuous for x,y\ne 0 since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at x,y=0 since it is trivial to check that $

\displaystyle 0=F(0\times y_0)=F(x_0\times 0)=\lim_{x\to 0}F(x\times y_0)=\lim_{y\to 0}F(x_0\times y)

b) Evidently

\displaystyle g(x)=F(x\times x)=\begin{cases}\frac{1}{2}\quad\text{if}\quad x\times x\ne 0\\ 0\quad\text{if}\quad x\times x=0\end{cases}

c) This clearly proves that F(x\times y) is not continuous with \mathbb{R}^2 is not continuous since if \Delta is the diangonal we have that

\displaystyle \lim_{(x,y)\in\Delta\to (0,0)}F(x\times y)=\frac{1}{2}\ne F(0\times 0)

and so in particular

\displaystyle \lim _{(x,y)\to(0,0)}F(x\times y)\ne F(0\times 0)


Problem: Let A\subseteq X; let f:A\to X be continuous and let Y be Hausdorff. Prove that if f may be extended to a continuous function \overset{\sim}{f}:\overline{A}\to Y, then \overset{\sim}{f} is uniquely determined by f.

Proof: I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways

Way 1(fun way!):

Lemma: Let X be any topological space and Y a Hausdorff space. Suppose that \varphi,\psi:X\to Y are continuous and define A(\varphi,\psi)=\left\{x\in X:\varphi(x)=\psi(x)\right\}. Then, A(\varphi,\psi) is closed in X

Proof: Note that \varphi\oplus\psi:X\to Y\times Y:x\mapsto (\varphi(x),\psi(x)) is clearly continuous since \pi_1\circ(\varphi\oplus\psi)=\varphi and \pi_2\circ(\varphi\oplus\psi)=\psi. It is trivial then to check that \displaystyle A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta_Y) and since Y is Hausdorff we have that \Delta_Y\subseteq Y\times Y is closed and the conclusion follows. \blacksquare

From this we note that if \varphi,\psi agree on D\subseteq X such that \overline{D}=X we have that

X\supseteq A(\varphi,\psi)=\overline{A(\varphi,\psi)}\supseteq\overline{D}=X

From where it follows that A(\varphi,\psi)=X and so \varphi=\psi. So, thinking of \overline{A} as a subspace of X we see that \text{cl}_{\overline{A}}\text{ }A=Y\cap\text{cl}_{X}\text{ }A=\overline{A} and thus clearly A is dense in \overline{A}. So, the conclusion readily follows by noting that if \overset{\sim}{f_1},\overset{\sim}{f_2} are two continuous extensions then by definition A\left(\overset{\sim}{f_1},\overset{\sim}{f_2}\right)\supseteq A.

Way 2(unfun way): Let \overset{\sim}{f_1},\overset{\sim}{f_2} be two extensions of f and suppose there is some x\in\overline{A}-A(\varphi,\psi). Clearly x\notin A and thus x is a limit point of A. So, by assumption \overset{\sim}{f_1}(x)\ne\overset{\sim}{f_2}(x) and so using the Hausdorffness of Y we may find disjoint neighborhoods U,V of them respectively. Thus, \overset{\sim}{f_1}^{-1}(U),\overset{\sim}{f_2}^{-1}(V) are neighborhoods of x in X. Thus, \overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V) is a neighborhood of x. But, clearly there can be no y\in A\cap\left(\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)\right) otherwise \overset{\sim}{f_1}(y)=\overset{\sim}{f_2}(y)\in U\cap V. It follows that \overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V) is a neighborhood of x disjoint from A which contradicts the density of A in \overline{A}.  The conclusion follow. \blacksquare

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May 28, 2010 - Posted by | Fun Problems, Munkres, Topology, Uncategorized | , , , , ,


  1. A comment on your solution to #2: it appears that defining X=(-1,0)\cup(0,1) is unnecessary, since X=(-1,0) would work just as well. I suspect that your example is a holdover from an early draft of your solution when you were likely trying to consider 0 as the limit point.

    At any rate, you also have two typos: f^{-1}(\{0\})=(-1,0) as opposed to f^{-1}\{(1\})=(-1,0), and you have an extra \cap in your third to last tex string.

    Comment by soiteroo | July 3, 2011 | Reply

    • Ah, thank you! You are one-hundred percent correct! It was a holdover from a previous draft. I had done all of these far before I posted them up here and was anxious to get them down, they are also smattered with problems from other texts by accident. I will fix that typo then, thanks!

      Comment by Alex Youcis | July 8, 2011 | Reply

  2. Also, though it’s essentially what you’re doing I think 9a is even more trivial when you recognize it as induction on the number of pasted-together sets in the pasting lemma.

    [I'm not trying to be disagreeable; I'm going through the same project as you and I've been comparing my solutions to yours.]

    Comment by soiteroo | July 3, 2011 | Reply

    • Friend,

      Thanks for the input! I see what you’re saying, and I agree that induction would work fine! I think it is more of a stylistic difference than anything.

      I don’t find you disagreeable in the slightest, I really enjoy when people point out flaws in my solutions/alternate (better) solutions! Math is a communal subject, and any feedback (good or bad) helps me learn.

      I wish you luck in your goal of finishing Munkres. I eventually got restless and moved on to other subjects, but I got fairly far (I just didn’t post them). If you have any specific questions about a problem feel free to e-mail me!

      Comment by Alex Youcis | July 8, 2011 | Reply

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